Miscellaneous Exercise : Proofs of Trigonometric Identities

1. Prove that $\dfrac{\tan \theta+\cot \theta}{\sec \theta+\operatorname{cosec} \theta}=\dfrac{1}{\sin \theta+\cos \theta}$





$\begin{aligned} &\dfrac{\tan \theta+\cot \theta}{\sec \theta+\csc \theta}\\\\ = &\dfrac{\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\cos \theta}+\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}}{\dfrac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}}\\\\ = &\dfrac{1}{\sin \theta \cos \theta} \times \dfrac{\sin \theta \cos \theta}{\sin \theta+\cos \theta}\\\\ = &\dfrac{1}{\sin \theta+ \cos \theta} \end{aligned}$



2. Prove that $\quad \sec x \csc x-\cot x=\tan x$.





$\begin{aligned} & \sec x \csc x-\cot x \\\\ = & \dfrac{1}{\cos x} \dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x} \\\\ = & \dfrac{1-\cos ^{2} x}{\sin x \cos x} \\\\ = & \dfrac{\sin ^{2} x}{\sin x \cos x} \\\\ = & \dfrac{\sin x}{\cos x} \\\\ = & \tan x \end{aligned}$



3. Prove that $\dfrac{1}{1-\cos x}+\dfrac{1}{1+\cos x}=2 \csc^{2} x$





$\begin{aligned} & \dfrac{1}{1-\cos x}+\dfrac{1}{1+\cos x} \\\\ = & \dfrac{1+\cos x+1-\cos x}{(1-\cos x)(1+\cos x)} \\\\ = & \dfrac{2}{1-\cos ^{2} x} \\\\ = & \dfrac{2}{\sin ^{2} x} \\\\ = & 2 \csc ^{2} x \end{aligned}$



4. Show that $\dfrac{\tan \theta+\cot \theta}{\csc \theta}=\sec \theta$





$\begin{aligned} &\dfrac{\tan \theta+\cot \theta}{\csc \theta}\\\\ = &\dfrac{\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}}{\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{1}{\sin \theta \cos \theta} \times \sin \theta \\\\ = &\sec \theta \end{aligned}$



5. Show that $\sqrt{\sec ^{2} \theta-1}+\sqrt{\csc^{2} \theta-1}=\sec \theta \csc \theta$





$\begin{aligned} & \sqrt{\sec ^{2} \theta-1}+\sqrt{\csc ^{2} \theta-1} \\\\ =& \sqrt{\tan ^{2} \theta}+\sqrt{\cot ^{2} \theta} \\\\ =& \tan \theta+\cot \theta \\\\ =& \dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta} \\\\ =& \dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta} \\\\ =& \dfrac{1}{\sin \theta \cos \theta} \\\\ =& \sec \theta \csc \theta \end{aligned}$



6. Prove that $\sec ^{2} x+\csc^{2} x=\sec ^{2} x \csc^{2} x$





$\begin{aligned} & \sec ^{2} x+\csc ^{2} x \\\\ =& \dfrac{1}{\cos ^{2} x}+\dfrac{1}{\sin ^{2} x} \\\\ =& \dfrac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} \\\\ =& \dfrac{1}{\sin ^{2} x \cos ^{2} x} \\\\ =& \dfrac{1}{\cos ^{2} x} \cdot \dfrac{1}{\sin ^{2} x}\\\\ =&\sec ^{2} x \csc ^{2} x \end{aligned}$



7. Show that $\dfrac{1}{\csc \theta-1}-\dfrac{1}{\csc \theta+1}=2 \tan ^{2} \theta$





$\begin{aligned} & \dfrac{1}{\csc \theta-1}-\dfrac{1}{\csc \theta+1} \\\\ =& \dfrac{(\csc \theta+1)-(\csc \theta-1)}{(\csc \theta-1)(\csc \theta+1)} \\\\ =& \dfrac{2}{\csc ^{2} \theta-1} \\\\ =& \dfrac{2}{\cot ^{2} \theta} \\\\ =& 2 \tan ^{2} \theta \end{aligned}$



8. Show that $\dfrac{\csc \theta}{\csc \theta-\sin \theta}=\sec ^{2} \theta$.





$\begin{aligned} &\dfrac{\csc \theta}{\csc \theta-\sin \theta} \\\\ =& \dfrac{\dfrac{1}{\sin \theta}}{\dfrac{1}{\sin \theta}-\sin \theta} \\\\ =& \dfrac{1}{\sin \theta} \times \dfrac{\sin \theta}{\cos ^{2} \theta}\\\\ =&\sec ^{2} \theta \end{aligned}$



9. Prove that $\dfrac{\cos x}{1+\tan x}-\dfrac{\sin x}{1+\cot x}=\cos x-\sin x$.





$\begin{aligned} &\dfrac{\cos x}{1+\tan x}-\dfrac{\sin x}{1+\cot x} \\\\ =&\dfrac{\cos x}{1+\dfrac{\sin x}{\cos x}}-\dfrac{\sin x}{1+\dfrac{\cos x}{\sin x}} \\\\ =&\dfrac{\cos x+\sin x}{\cos x}-\dfrac{\sin x}{\dfrac{\sin x+\cos u}{\sin x}}\\\\ =&\cos x\left(\dfrac{\cos x}{\cos x+\sin x}\right)-\sin x\left(\dfrac{\sin x}{\sin x+\cos x}\right)\\\\ =&\dfrac{\cos ^{2} x-\sin ^{2} x}{\cos x+\sin x}\\\\ =&\dfrac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x+\sin x}\\\\ =&\cos x-\sin x \end{aligned}$



10. Show that $\csc \theta-\sin \theta=\cot \theta \cos \theta$.





$\begin{aligned} & \csc \theta-\sin \theta \\\\ =& \dfrac{1}{\sin \theta}-\sin \theta \\\\ =& \dfrac{1-\sin ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{\cos ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{\cos \theta}{\sin \theta} \cos \theta \\\\ =& \cot \theta \cos \theta \end{aligned}$



11. Show that $\dfrac{\tan ^{2} \theta+\sin ^{2} \theta}{\cos \theta+\sec \theta}=\tan \theta \sin \theta$.





$\begin{aligned} & \dfrac{\tan ^{2} \theta+\sin ^{2} \theta}{\cos \theta+\sec \theta} \\\\ =& \dfrac{\dfrac{\sin ^{2} \theta}{\cos ^{2} \theta}+\sin ^{2} \theta}{\cos \theta+\dfrac{1}{\cos \theta}} \\\\ =& \dfrac{\sin ^{2} \theta\left(1+\cos ^{2} \theta\right)}{\cos ^{2} \theta} \times \dfrac{\cos \theta}{\cos ^{2} \theta+T}\\\\ =&\dfrac{\sin ^{2} \theta}{\cos \theta} \\\\ =&\dfrac{\sin \theta}{\cos \theta} \sin \theta \\\\ =&\tan \theta \sin \theta \end{aligned}$



12. Prove that $\sin x(\cot x+\tan x)=\sec x$





$\begin{aligned} & \sin x(\cot x+\tan x) \\ =& \sin x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right) \\\\ =& \sin x\left(\dfrac{\cos 2 x+\sin ^{2} x}{\sin x \cos x}\right) \\\\ =& \dfrac{1}{\cos x} \\\\ =& \sec x \end{aligned}$



13. Show that $\dfrac{\sec \theta}{\cot \theta+\tan \theta}=\sin \theta$.





$\begin{aligned} &\dfrac{\sec \theta}{\cot \theta+\tan \theta}\\\\ =&\dfrac{\dfrac{1}{\cos \theta}}{\dfrac{\cos \theta}{\sin \theta}+\dfrac{\sin \theta}{\cos \theta}}\\\\ =&\dfrac{\dfrac{1}{\cos \theta}}{\dfrac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}}\\\\ =&\dfrac{1}{\cos \theta} \times \dfrac{\sin \theta \cos \theta}{1}\\\\ =&\sin \theta \end{aligned}$



14. Show that $\dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x}=2 \csc x$.





$\begin{aligned} & \dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x}{1+\cos x} \times \dfrac{1-\cos x}{1-\cos x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x(1-\cos x)}{1-\cos ^{2} x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x(1-\cos x)}{\sin ^{2} x}+\dfrac{1+\cos x}{\sin x}\\\\ =&\dfrac{1-\cos x}{\sin x}+\dfrac{1+\cos x}{\sin x} \\\\ =&\dfrac{1-\cos x+1+\cos x}{\sin x} \\\\ =&\dfrac{2}{\sin x} \\\\ =&2 \csc x \end{aligned}$



15. Show that $\dfrac{(1-\sin A)(1+\sin A)}{\sin A \cos A}=\cot A$.





$\begin{aligned} & \dfrac{(1-\sin A)(1+\sin A)}{\sin A \cos A} \\\\ =& \dfrac{1-\sin ^{2} A}{\sin A \cos A} \\\\ =& \dfrac{\cos ^{2} A}{\sin A \cos A} \\\\ =& \dfrac{\cos A}{\sin A}\\\\ =&\cot A \end{aligned}$



16. Show that $\cos \theta \cot \theta+\sin \theta=\csc \theta$.





$\begin{aligned} & \cos \theta \cot \theta+\sin \theta \\\\ =& \cos \theta \cdot \dfrac{\cos \theta}{\sin \theta}+\sin \theta \\\\ =& \dfrac{\cos ^{2} \theta}{\sin \theta}+\sin \theta \\\\ =& \dfrac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{1}{\sin \theta} \\\\ =& \csc \theta \end{aligned}$



17. Show that $2 \cos x \cot x+1=\cot x+2 \cos x$ can be written in the form $(a \cos x-b)(\cos x-\sin x)=0$, where $a$ and $b$ are constants to be found.





$\begin{aligned} &\quad 2 \cos x \cot x+1=\cot x+2 \cos x \\\\ &\quad 2 \cos x \dfrac{\cos x}{\sin x}+1=\dfrac{\cos x}{\sin x}+2 \cos x \\\\ &\quad \text { Multiplying both sides with}\ \sin x, \\\\ &\quad 2 \cos ^{2} x+\sin x=\cos x+2 \sin x \cos x \\\\ &\quad 2 \cos ^{2} x-\cos x+\sin x-2 \sin x \cos x=0 \\\\ &\quad\cos x(2 \cos x-1)+\sin x(1-2 \cos x)=0\\\\ &\quad\cos x(2 \cos x-1)-\sin x(2 \cos x-1)=0 \\\\ &\quad(2 \cos x-1)(\cos x-\sin x)=0 \\\\ &\therefore(a \cos x-b)(\cos x-\sin x)=(2 \cos x-1)(\cos x-\sin x) \\\\ &\therefore a=2\quad \text { and }\quad b=1 \end{aligned}$

Composition of Functions : Exercise (4.8) - Solutions

Key Points 

• Some pairs of functions cannot be composed. Some pairs of functions can be composed only for certain values of $x$.
  
  
• The domain of a composite function is either the same as the domain of the first function, or else lies inside it.
  
  
• The range of a composite function is either the same as the range of the second function, or else lies inside it.

1.           Functions $f$ and $g$ are given by $f(x) = 2x + 1$ and $g(x) = 3x$.

              (a)   Calculate $(g\circ f)(1)$ and $(g\circ f)(3)$.

              (b)   Find the formula of $(g\circ f)$ and check the above images. State the domainof $(g\circ f)$.

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$\begin{array}{*{20}{l}} {f(x)=2x+1,\;g(x)=3x} \\ {} \\ {(g\circ f)(1)\;=g\left( {f(1)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=g\left( {2\times 1+1} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=g(3)} \\ {\;\;\;\;\;\;\;\;\;\;=3\times 3} \\ {\;\;\;\;\;\;\;\;\;\;=9} \\ {} \\ {(g\circ f)(3)\;=g\left( {f(3)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=g\left( {2\times 3+1} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=g(7)} \\ {\;\;\;\;\;\;\;\;\;\;=3\times 7} \\ {\;\;\;\;\;\;\;\;\;\;=21} \\ {} \\ {(g\circ f)(x)\;=g\left( {f(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=g\left( {2x+1} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=3\left( {2x+1} \right)} \\ {} \\ {f\;\text{is defined for all real numbers of}\;x.} \\ \begin{array}{l}\therefore \;\;\operatorname{dom}(f)=\mathbb{R}\\3\left( {2x+1} \right)\ \text{is defined for all real numbers of}\;x.\ \end{array} \\ {\therefore \;\;\operatorname{dom}(g\circ f)=\mathbb{R}} \end{array}$

2.           The functions $f$ and $g$ are given by $f(x) = x + 2$ and $g(x) = x^2$.

              (a)   Find the formulae for $(g\circ f)$,$(g\circ g)$, $(f\circ g)$, $(f\circ f)$, and their domains.

              (b)   Find $(g\circ f)(−1)$ and $(g\circ f)(2)$.

              (c)   Find $(f\circ g)(−1)$ and $(f\circ g)(2)$.

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$\begin{array}{*{20}{l}} {f(x)=x+2,\;g(x)={{x}^{2}}} \\ {} \\ {\left( {g\circ f} \right)(x)=g\left( {f(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=g\left( {x+2} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;={{{\left( {x+2} \right)}}^{2}}} \\ {} \\ {\left( {g\circ g} \right)(x)=g\left( {g(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=g\left( {{{x}^{2}}} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;={{{\left( {{{x}^{2}}} \right)}}^{2}}} \\ {\;\;\;\;\;\;\;\;\;\;\;={{x}^{4}}} \\ {} \\ {\left( {f\circ g} \right)(x)=f\left( {g(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=f\left( {{{x}^{2}}} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;={{x}^{2}}+2} \\ {} \\ {\left( {f\circ f} \right)(x)=f\left( {f(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=f\left( {x+2} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=x+2+2} \\ {\;\;\;\;\;\;\;\;\;\;\;=x+4} \\ {} \\ {\text{Both }\;f(x)\;\text{and }g(x)\;\text{are defined for all real numbers of}\;x.} \\ {} \\ {{{{\left( {x+2} \right)}}^{2}},\ {{x}^{4}},\ {{x}^{2}}+2\ \text{and}\ x+4\ \text{are defined for all real numbers of}\;x.\ } \\ {} \\ {\operatorname{dom}\left( {g\circ f} \right)=\operatorname{dom}(f)=\mathbb{R}} \\ {\operatorname{dom}\left( {g\circ g} \right)=\operatorname{dom}(g)=\mathbb{R}} \\ {\operatorname{dom}\left( {f\circ g} \right)=\operatorname{dom}(g)=\mathbb{R}} \\ {\operatorname{dom}\left( {f\circ f} \right)=\operatorname{dom}(f)=\mathbb{R}} \end{array}$

3.           Find the formulae for composite functions $(f\circ g)$, $(g\circ f)$ and their domains in each case.

              $\begin{array}{ll} \text{(a)}\ \ f(x)=x+1, & g(x)=2 x^{2}-x+3\\\\ \text{(b)}\ \ f(x)=x^{2}-1, & g(x)=3 x+1\\\\ \text{(c)}\ \ f(x)=-x, & g(x)=x\\\\ \text{(d)}\ \ f(x)=x^{2}, & g(x)=\sqrt{x} \end{array}$

Solutions

              $\text{(a)}\ \ f(x)=x+1, \ \ \ \ \ g(x)=2 x^{2}-x+3$

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$\begin{array}{l}\left( {f\circ g} \right)(x)=f\left( {g(x)} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =f\left( {2{{x}^{2}}-x+3} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =2{{x}^{2}}-x+3+1\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =2{{x}^{2}}-x+4\\\\\left( {g\circ f} \right)(x)=g\left( {f(x)} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =g\left( {x+1} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =2{{\left( {x+1} \right)}^{2}}-\left( {x+1} \right)+3\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =2{{x}^{2}}+3x+4\\\\\text{Both }\;f(x)\;\text{and }g(x)\;\text{are defined for all real numbers of}\;x.\\\\\operatorname{dom}(f)=\mathbb{R}\;\text{and}\;\operatorname{dom}(g)=\mathbb{R}\\\\\text{Similarly, both }2{{x}^{2}}-x+4\;\text{and }2{{x}^{2}}+3x+4\;\text{are }\\\text{defined for all real numbers of}\;x.\\\\\operatorname{dom}\left( {f\circ g} \right)=\mathbb{R}\;\text{and}\;\operatorname{dom}\left( {g\circ f} \right)=\mathbb{R}\end{array}$

              $\text{(b)}\ \ f(x)=x^{2}-1, \ \ \ \ \ g(x)=3 x+1$

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$\begin{array}{*{20}{l}} {\left( {f\circ g} \right)(x)=f\left( {g(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=f\left( {3x+1} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;={{{\left( {3x+1} \right)}}^{2}}-1} \\ {\;\;\;\;\;\;\;\;\;\;\;=9{{x}^{2}}+6x} \\ {} \\ {\left( {g\circ f} \right)(x)=g\left( {f(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=g\left( {{{x}^{2}}-1} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=3\left( {{{x}^{2}}-1} \right)+1} \\ {\;\;\;\;\;\;\;\;\;\;\;=3{{x}^{2}}-2} \\ {} \\ \begin{array}{l}\text{Both }\;f(x)\;\text{and }g(x)\;\text{are defined for all real numbers of}\;x.\\\\\therefore \ \ \operatorname{dom}(f)=\mathbb{R}\;\text{and}\;\operatorname{dom}(g)=\mathbb{R}\\\\\text{Similarly both }9{{x}^{2}}+6x\;\text{and }3{{x}^{2}}-2\;\text{are defined for }\\\text{all real numbers of}\;x.\end{array} \\ {} \\ {\therefore \ \ \operatorname{dom}\left( {f\circ g} \right)=\mathbb{R}\ \ \text{and}\ \operatorname{dom}\left( {g\circ f} \right)=\mathbb{R}} \end{array}$

              $\text{(c)}\ \ f(x)=-x, \ \ \ \ \ g(x)=x$

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$ \begin{array}{l}\left( {f\circ g} \right)(x)=f\left( {g(x)} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ =f\left( x \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ =-x\\\\\left( {g\circ f} \right)(x)=g\left( {f(x)} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ =g\left( {-x} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ =-x\\\\\text{Both }\;f(x)\;\text{and }g(x)\;\text{are defined for all real numbers of}\;x.\\\\\operatorname{dom}(f)=\mathbb{R}\;\text{and}\;\operatorname{dom}(g)=\mathbb{R}\\\\\text{Also}-x\ \text{is}\ \text{defined for all real numbers of}\;x.\\\\\operatorname{dom}\left( {f\circ g} \right)=\mathbb{R}\ \text{and}\ \operatorname{dom}\left( {g\circ f} \right)=\mathbb{R}\end{array}$

              $\text{(d)}\ \ f(x)=x^2, \ \ \ \ \ g(x)=\sqrt{x}$

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$\begin{array}{*{20}{l}} {\left( {f\circ g} \right)(x)=f\left( {g(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;=f\left( {\sqrt{x}} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;={{{\left( {\sqrt{x}} \right)}}^{2}}} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;=x} \\ {} \\ {\left( {g\circ f} \right)(x)=g\left( {f(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;=g\left( {{{x}^{2}}} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;=\sqrt{{{{x}^{2}}}}} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;=x} \\ {} \\ {f(x)\;\text{is defined for all real numbers of}\;x.} \\ {} \\ {\operatorname{dom}(f)=\mathbb{R}\;} \\ {} \\ {g(x)\;\text{is defined for }x\ge 0,\;x\in \mathbb{R}} \\ {} \\ {\operatorname{dom}(g)=\left\{ {x\;|\;x\ge 0,\;x\in \mathbb{R}} \right\}} \\ {} \\ \begin{array}{l}x\ \text{is defined for all real numbers of}\;x\ \text{but}\\\sqrt{x}\ \text{is defined for}\ x\ge 0.\ \\\\\operatorname{dom}\left( {f\circ g} \right)=\operatorname{dom}(g)=\left\{ {x\;|\;x\ge 0,\;x\in \mathbb{R}} \right\}\\\end{array} \\ {\operatorname{dom}\left( {g\circ f} \right)=\operatorname{dom}(f)=\mathbb{R}} \end{array}$

4.           A function $f$ is given by $f(x)=x+1$. Find the function $g:\mathbb{R}\to \mathbb{R}$ in each of the following:

              $\begin{array}{l} \text{(a)}\ \ \left( {g\circ f} \right)(x) = x^2+5x+5\\\\ \text{(b)}\ \ \left( {f\circ g} \right)(x) = x^2+5x+5 \end{array}$

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$\begin{array}{l} \ \ \ \ \ \ \ \ f(x)=x+1,\ \ g:\mathbb{R}\to \mathbb{R}\\\\ \text{(a)}\;\;\left( {g\circ f} \right)(x)={{x}^{2}}+5x+5\\\\ \ \ \ \ \ \ \ \ g\left( {f(x)} \right)={{x}^{2}}+5x+5\\\\ \ \ \ \ \ \ \ \ g\left( {x+1} \right)={{x}^{2}}+2x+1+3x+3+1\\\\ \ \ \ \ \ \ \ \ g\left( {x+1} \right)={{(x+1)}^{2}}+3(x+1)+1\\\\ \therefore \ \ \ g(x)={{x}^{2}}+3x+1\\\\\\ \text{(b)}\;\;\left( {f\circ g} \right)(x)={{x}^{2}}+5x+5\\\\ \ \ \ \ \ \ \ \ f\left( {g(x)} \right)={{x}^{2}}+5x+5\\\\ \ \ \ \ \ \ \ \ g(x)+1={{x}^{2}}+5x+5\\\\ \therefore \ \ \ g(x)={{x}^{2}}+5x+4 \end{array}$

5.           If $g:\mathbb{R}\to \mathbb{R}$ is given by $g(x)=x^2+3$, find the function $g$ such that

              $\begin{array}{l} \text{(a)}\ \ \left( {f\circ g} \right)(x)= 4x^2+3\\\\ \text{(b)}\ \ \left( {g\circ f} \right)(x)= 4x^2+3 \end{array}$

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$\begin{array}{l}\ \ \ \ \ \ \ g:\mathbb{R}\to \mathbb{R},\ \ \ g(x)={{x}^{2}}+3\\\\\text{(a)}\;\;\left( {f\circ g} \right)(x)=4{{x}^{2}}+3\\\\\ \ \ \ \ \ \ \ f\left( {g(x)} \right)=4{{x}^{2}}+3\\\\\ \ \ \ \ \ \ \ f\left( {{{x}^{2}}+3} \right)=4{{x}^{2}}+12-9\\\\\ \ \ \ \ \ \ \ f\left( {{{x}^{2}}+3} \right)=4\left( {{{x}^{2}}+3} \right)-9\\\\\therefore \ \ \ \ \ \ \ f\left( x \right)=4x-9\\\\\\\text{(b)}\;\;\left( {g\circ f} \right)(x)=4{{x}^{2}}+3\\\\\ \ \ \ \ \ \ \ g\left( {f(x)} \right)=4{{x}^{2}}+3\\\\\ \ \ \ \ \ \ \ {{\left( {f(x)} \right)}^{2}}+3=4{{x}^{2}}+3\\\\\ \ \ \ \ \ \ \ \ {{\left( {f(x)} \right)}^{2}}=4{{x}^{2}}\\\\\therefore \ \ \ \ \ \ \ f(x)=\pm \ 2x\end{array}$

6.           Functions $f$ and $g$ are given by $f(x)=px-2$ where $p$ is a constant and $g(x)=4x+3$. Find the value of $p$ for which $\left( {f\circ g} \right)(x) = \left( {g\circ f} \right)(x)$.

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$\begin{array}{l}\ \ \ \ f(x)=px-2,\ \ \ g(x)=4x+3\\\\\ \ \ \;\left( {f\circ g} \right)(x)=\left( {g\circ f} \right)(x)\\\\\ \ \ \ f\left( {g(x)} \right)=g\left( {f(x)} \right)\\\\\ \ \ \ f\left( {4x+3} \right)=g\left( {px-2} \right)\\\\\ \ \ \ p\left( {4x+3} \right)-2=4\left( {px-2} \right)+3\\\\\ \ \ \ 4px+3p-2=4px-8+3\\\\\ \ \ \ 3p-2=-5\\\\\ \ \ \ p=-1\end{array}$

7.           Let $f$ and $g$ be functions given by $f(x) = 3x− 1$ and $g(x) = x+7$. Find the formulae of $f^{-1}\circ g,\ g^{-1}\circ f $ and state their domains. What are the values of $\left(f^{-1}\circ g\right)(3)$ and $\left(g\circ f^{-1}\right)(2)$.

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$\begin{array}{l}\ \ \ \ f(x)=3x-1,\ \ \ g(x)=x+7\\\\\ \ \ \;\text{Let}\ {{f}^{{-1}}}(x)=y\\\\\ \ \ \ f\left( y \right)=x\\\\\ \ \ \ 3y-1=x\\\\\therefore \ \ y=\displaystyle \frac{{x+1}}{3}\\\\\therefore \ \ {{f}^{{-1}}}(x)=\displaystyle \frac{{x+1}}{3}\\\\\therefore \ \ \left( {{{f}^{{-1}}}\circ g} \right)(x)={{f}^{{-1}}}\left( {g(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( {x+7} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{x+7+1}}{3}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{x+8}}{3}\\\\\ \ \ \ \text{Both }f(x)\ \text{and }g(x)\ \text{are defined for all real numbers of }x.\\\\\therefore \ \operatorname{dom}(f)=\mathbb{R},\ \operatorname{dom}(g)=\mathbb{R}\\\\\ \ \displaystyle \frac{{x+8}}{3}\ \text{ defined for all real numbers of }x.\\\\\therefore \ \operatorname{dom}({{f}^{{-1}}}\circ g)=\mathbb{R}\\\\\ \ \ \ \text{Let}\ {{g}^{{-1}}}(x)=z\\\\\ \ \ \ g(z)=x\\\\\ \ \ \ z+7=x\\\\\ \ \ \ z=x-7\\\\\therefore \ \ {{g}^{{-1}}}(x)=x-7\\\\\therefore \ \ \left( {{{g}^{{-1}}}\circ f} \right)(x)={{g}^{{-1}}}\left( {f(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{g}^{{-1}}}\left( {3x-1} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3x-1-7\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3x-8\\\\\ \ \ \ \text{Both }f(x)\ \text{and }g(x)\ \text{are defined for all real numbers of }x.\\\\\therefore \ \operatorname{dom}(f)=\mathbb{R},\ \operatorname{dom}(g)=\mathbb{R}\\\\\ \ 3x-8\ \text{ defined for all real numbers of }x.\\\\\therefore \ \operatorname{dom}({{g}^{{-1}}}\circ f)=\mathbb{R}\\\\\therefore \ \left( {{{f}^{{-1}}}\circ g} \right)(3)=\displaystyle \frac{{3+8}}{3}=\displaystyle \frac{{11}}{3}\\\\\ \ \ \ \ \left( {g\circ {{f}^{{-1}}}} \right)(2)=g\left( {{{f}^{{-1}}}(2)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =g\left( {\displaystyle \frac{{2+1}}{3}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =g\left( 1 \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+7\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =8\end{array}$

8.           Let the functions $f$ and $g$ be given by $f(x) = 2x−1$ and $g(x) =\displaystyle \frac{2x+3}{x-1}$.

Find the composite function $f\circ g$.

Find the inverse function $f^{-1}$ and $g^{-1}$.

Evaluate $\left(f\circ g^{-1}\right)(1)$ and $\left(f^{-1}\circ g^{-1}\right)(1)$.

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$\begin{array}{l}\ f(x)=2x-1\\\ g(x)=\displaystyle \frac{{2x+3}}{{x-1}}\\\\\ \left( {f\circ g} \right)(x)=f\left( {g(x)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\displaystyle \frac{{2x+3}}{{x-1}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\displaystyle \frac{{2x+3}}{{x-1}}} \right)-1\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{4x+6-x+1}}{{x-1}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{3x+7}}{{x-1}}\\\\\ \text{Let}\ {{f}^{{-1}}}(x)=y,\ \text{then}\\\ f(y)=x\\\ 2y-1=x\\\ y=\displaystyle \frac{{x+1}}{2}\\\ {{f}^{{-1}}}(x)=\displaystyle \frac{{x+1}}{2}\\\\\text{Let}\ {{g}^{{-1}}}(x)=z,\ \text{then}\\\ g(z)=x\\\ \displaystyle \frac{{2z+3}}{{z-1}}=x\\\ 2z+3=xz-x\\\ xz-2z=x+3\\\ z(x-2)=x+3\\\ z=\displaystyle \frac{{x+3}}{{x-2}}\\\ {{g}^{{-1}}}(x)=\displaystyle \frac{{x+3}}{{x-2}}\\\\\left( {f\circ {{g}^{{-1}}}} \right)(1)=f\left( {{{g}^{{-1}}}(1)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\displaystyle \frac{{1+3}}{{1-2}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {-4} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\left( {-4} \right)-1\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-9\\\\\left( {{{f}^{{-1}}}\circ {{g}^{{-1}}}} \right)(1)={{f}^{{-1}}}\left( {{{g}^{{-1}}}(1)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( {\displaystyle \frac{{1+3}}{{1-2}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( {-4} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-4+1}}{2}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{3}{2}\end{array}$

9.           Let the functions $f$, $g$ and $h$ be 𝑓(𝑥) = $f(x) = x−2$, $g(x) = x^3$ and $h(x) = 4x$. Show that $\left(h\circ g\right)\circ f= h\circ \left(g\circ f\right)$.

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$ \begin{array}{l}\ f(x)=x-2\\\ g(x)={{x}^{3}}\\\ h(x)=4x\\\\\ \left( {h\circ g} \right)(x)=h\left( {g(x)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =h\left( {{{x}^{3}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{x}^{3}}\\\ \\\ \left( {\left( {h\circ g} \right)\circ f} \right)(x)=\left( {h\circ g} \right)\left( {f(x)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {h\circ g} \right)\left( {x-2} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{\left( {x-2} \right)}^{3}}\\\\\ \left( {g\circ f} \right)(x)=g\left( {f(x)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =g\left( {x-2} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\left( {x-2} \right)}^{3}}\\\ \\\ \left( {h\circ \left( {g\circ f} \right)} \right)(x)=h\left( {\left( {g\circ f} \right)(x)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =h\left( {{{{\left( {x-2} \right)}}^{3}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{\left( {x-2} \right)}^{3}}\\\\\ \therefore \ \ \left( {\left( {h\circ g} \right)\circ f} \right)(x)=\left( {h\circ \left( {g\circ f} \right)} \right)(x)\\\ \therefore \ \ \left( {h\circ g} \right)\circ f=h\circ \left( {g\circ f} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

Logarithms : Exercise (3.3) Solutions

$\begin{array}{|l|l|l|} \hline {\ \ \ \ \ \ \ \ \ \ \text { Properties }} & {\ \ \ \ \ \ \ \text { For Exponents }} & {\ \ \ \ \ \ \ \ \ \ \ \ \ \text { For Logarithms }} \\ \hline \text { One-to-one Property } & \text { If } b^{x}=b^{y}, \text { then } x=y & \text { If } \log _{b} M=\log _{b} N, \text { then } M=N \\ \hline \text { Product Property } & b^{x} \cdot b^{y}=b^{x+y} & \log _{b}(M N)=\log _{b} M+\log _{b} N \\ \hline \text { Quotient Property } & \displaystyle\frac{b^{x}}{b^{y}}=b^{x-y} & \log _{b} \displaystyle\frac{M}{N}=\log _{b} M-\log _{b} N \\ \hline \text { Power Property } & \left(b^{x}\right)^{y}=b^{x y} & \log _{b} N^{p}=p \log _{b} N \\ \hline \end{array}$

1.           Replace $\square$ with the appropriate number.

              $\begin{array}{l} \text{(a)}\ \ \log _{3} 24=\log _{3} 6+\log _{3} \square\\ \text{(b)}\ \ \log _{5} 24=\log _{5} 60+\log _{5} \square\\ \text{(c)}\ \ \log _{2} \square=3 \log _{2} 3\\ \text{(d)}\ \ \log _{10} 9=\square \log _{10} 3\\ \text{(e)}\ \ \log _{8} 5=\log _{8} \square-\log _{8} 11 \end{array}$

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$\begin{array}{l} \text{(a)}\ \ 4\\ \text{(b)}\ \ \displaystyle\frac{2}{5}\\ \text{(c)}\ \ 27\\ \text{(d)}\ \ 2\\ \text{(e)}\ \ 55 \end{array}$

2.           Write each expression as a single logarithm.

              $\begin{array}{l} \text{(a)}\ \ \log _{b} 20+\log _{b} 57-\log _{b} 241\\ \text{(b)}\ \ 3 \log _{b} 8-\displaystyle\frac{1}{2} \log _{b} 12\\ \text{(c)}\ \ \log _{b} x-2 \log _{b} y-\log _{b} a\\ \text{(d)}\ \ \log _{2} 3+\log _{4} 15 \end{array}$

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$\begin{array}{l}\text{(a)}\;\;\ \ {{\log }_{b}}20+{{\log }_{b}}57-{{\log }_{b}}241\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{20\times 57}}{{241}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{1140}}{{241}}\\\\ \text{(b)}\;\;\ \ 3{{\log }_{b}}8-\displaystyle \frac{1}{2}{{\log }_{b}}12\\ \ \ \ \ =\ {{\log }_{b}}{{8}^{3}}-{{\log }_{b}}{{12}^{{\displaystyle \frac{1}{2}}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{8\times 8\times 8}}{{\sqrt{{12}}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{8\times 8\times 8}}{{2\sqrt{3}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{256}}{{\sqrt{3}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{256\sqrt{3}}}{3}\\\\ \text{(c)}\;\;\ \ {{\log }_{b}}x-2{{\log }_{b}}y-{{\log }_{b}}a\\ \ \ \ \ =\ {{\log }_{b}}x-{{\log }_{b}}{{y}^{2}}-{{\log }_{b}}a\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{x}{{a{{y}^{2}}}}\\\\ \text{(d)}\;\;\ \ {{\log }_{2}}3+{{\log }_{4}}15\\ \ \ \ \ \text{Let}\ {{\log }_{4}}15=x,\ \text{then}\\ \ \ \ \ 15={{4}^{x}}\\\ \ \ \ 15={{2}^{2}}^{x}\\ \ \ \ \ \therefore \ \ 2x={{\log }_{2}}15\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{1}{2}{{\log }_{2}}15\\ \ \ \ \ \therefore \ \ {{\log }_{4}}15={{\log }_{2}}\sqrt{{15}}\\ \ \ \ \ \therefore \ \ {{\log }_{2}}3+{{\log }_{4}}15\\ \ \ \ \ =\ \ {{\log }_{2}}3+{{\log }_{2}}\sqrt{{15}}\\ \ \ \ \ =\ \ {{\log }_{2}}3\sqrt{{15}} \end{array}$

3.           Write each expression in terms of $\log _{b} 2, \log _{b} 3$ and $\log _{b} 5$.

              $\begin{array}{l} \text{(a)}\ \ \log _{b} 8\\ \text{(b)}\ \ \log _{b} 15\\ \text{(c)}\ \ \log _{b} 270\\ \text{(d)}\ \ \log _{b} \displaystyle\frac{27 \sqrt[3]{5}}{16}\\ \text{(e)}\ \ \log _{b} \displaystyle\frac{216}{\sqrt[3]{32}}\\ \text{(f)}\ \ \log _{b}(648 \sqrt{125})\\ \end{array}$

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$\begin{array}{l} \text{(a)}\;\;{{\log }_{b}}8={{\log }_{b}}{{2}^{3}}=3{{\log }_{b}}2\\\\ \text{(b)}\;\;{{\log }_{b}}15={{\log }_{b}}(3\times 5)={{\log }_{b}}3+{{\log }_{b}}5\\\\ \text{(c)}\;\;{{\log }_{b}}270={{\log }_{b}}(2\times {{3}^{3}}\times 5)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}2+{{\log }_{b}}{{3}^{3}}+{{\log }_{b}}5\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}2+3{{\log }_{b}}3+{{\log }_{b}}5\\\\ \text{(d)}\;\;{{\log }_{b}}\displaystyle \frac{{27\sqrt[3]{5}}}{{16}}={{\log }_{b}}\displaystyle \frac{{{{3}^{3}}\times {{5}^{{\frac{1}{3}}}}}}{{{{2}^{4}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}{{3}^{3}}+{{\log }_{b}}{{5}^{{\frac{1}{3}}}}-{{\log }_{b}}{{2}^{4}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3{{\log }_{b}}3+\displaystyle \frac{1}{3}{{\log }_{b}}5-4{{\log }_{b}}2\\\\ \text{(e)}\;\;{{\log }_{b}}\displaystyle \frac{{216}}{{\sqrt[3]{{32}}}}={{\log }_{b}}\displaystyle \frac{{{{2}^{3}}\times {{3}^{3}}}}{{{{2}^{{\frac{5}{3}}}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}\left( {{{2}^{{^{{\frac{4}{3}}}}}}\times {{3}^{3}}} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{4}{3}{{\log }_{b}}2+3{{\log }_{b}}3\\\\ \text{(f)}\;\;{{\log }_{b}}(648\sqrt{{125}})={{\log }_{b}}({{2}^{3}}\times {{3}^{4}}\times {{5}^{{\frac{3}{2}}}})\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}{{2}^{3}}+{{\log }_{b}}{{3}^{4}}+{{\log }_{b}}{{5}^{{\frac{3}{2}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3{{\log }_{b}}2+4{{\log }_{b}}3+\displaystyle\frac{3}{2}{{\log }_{b}}5 \end{array}$

4.           Evaluate each expression.

              $\begin{array}{l} \text{(a)}\ \ \log _{2} 128\\ \text{(b)}\ \ \log _{3} 81^{4}\\ \text{(c)}\ \ \log _{\frac{1}{2}} 8\\ \text{(d)}\ \ \log _{8} 2\\ \text{(e)}\ \ \log _{3} \displaystyle\frac{\sqrt{3}}{81}\\ \text{(f)}\ \ \displaystyle\frac{\log _{3} \sqrt{3}}{\log _{3} 81}\\ \text{(g)}\ \ \displaystyle\frac{\log _{2} 25}{\log _{2} 5}\\ \text{(h)}\ \ \log _{4} 8 \end{array}$

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$\begin{array}{l} \text{(a)}\;\;{{\log }_{2}}128={{\log }_{2}}{{2}^{7}}=7\\\\ \text{(b)}\;\;{{\log }_{3}}{{81}^{4}}={{\log }_{3}}{{\left( {{{3}^{4}}} \right)}^{4}}={{\log }_{3}}{{3}^{{16}}}=16\\\\ \text{(c)}\;\;{{\log }_{{\frac{1}{2}}}}8={{\log }_{{\frac{1}{2}}}}{{2}^{3}}={{\log }_{{\frac{1}{2}}}}{{\left( {\displaystyle\frac{1}{2}} \right)}^{{-3}}}=-3\\\\ \text{(d)}\;\;{{\log }_{8}}2={{\log }_{8}}{{8}^{{\frac{1}{3}}}}=\displaystyle \frac{1}{3}\\\\ \text{(e)}\;\;{{\log }_{3}}\displaystyle \frac{{\sqrt{3}}}{{81}}={{\log }_{3}}\displaystyle \frac{{{{3}^{{\frac{1}{2}}}}}}{{{{3}^{4}}}}={{\log }_{3}}{{3}^{{-\frac{7}{2}}}}=-\displaystyle \frac{7}{2}\\\\ \text{(f)}\;\;\displaystyle \frac{{{{{\log }}_{3}}\sqrt{3}}}{{{{{\log }}_{3}}81}}=\displaystyle \frac{{{{{\log }}_{3}}{{3}^{{\frac{1}{2}}}}}}{{{{{\log }}_{3}}{{3}^{4}}}}=\displaystyle \frac{{\frac{1}{2}}}{4}=\displaystyle \frac{1}{8}\\\\ \text{(g)}\;\;\displaystyle \frac{{{{{\log }}_{2}}25}}{{{{{\log }}_{2}}5}}=\displaystyle \frac{{{{{\log }}_{2}}{{5}^{2}}}}{{{{{\log }}_{2}}5}}=\displaystyle \frac{{2{{{\log }}_{2}}5}}{{{{{\log }}_{2}}5}}=2\\\\ \text{(h)}\;\;{{\log }_{4}}8={{\log }_{4}}\sqrt{{64}}={{\log }_{4}}{{4}^{{\frac{3}{2}}}}=\displaystyle \frac{3}{2} \end{array}$

5.           Use $\log _{10} 2=0.3010$ and $\log _{10} 3=0.4771$ to evaluate each of the following expressions.

              $\begin{array}{lll} \text{(a)}\ \ \log _{10} 6 & \text{(b)}\ \ \log _{10} 1.5 & \text{(c)}\ \ \log _{10} \sqrt{3}\\\\ \text{(d)}\ \ \log _{10} 4 & \text{(e)}\ \ \log _{10} 4.5 & \text{(f)}\ \ \log _{10} 8\\\\ \text{(g)}\ \ \log _{10} 18 & \text{(h)}\ \ \log _{10} 5 \end{array}$

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$\begin{array}{l} {{\log }_{{10}}}2=0.3010,\ {{\log }_{{10}}}3=0.4771\\\\ \text{(a)}\;\;{{\log }_{{10}}}6=\;{{\log }_{{10}}}\left( {2\times 3} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}2+{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.3010+0.4771\\\ \ \ \ \ \ \ \ \ \ \ \ =0.7781\\\\ \text{(b)}\;\;{{\log }_{{10}}}1.5=\;{{\log }_{{10}}}\displaystyle \frac{3}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}3-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.4771-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.1761\\\\ \text{(c)}\;\;{{\log }_{{10}}}\sqrt{3}\ =\;{{\log }_{{10}}}{{3}^{{\frac{1}{2}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;\displaystyle \frac{1}{2}{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;\displaystyle \frac{1}{2}\times 0.4771\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;0.2386\\\\\text{(d)}\;\;{{\log }_{{10}}}4={{\log }_{{10}}}{{2}^{2}}\\ \ \ \ \ \ \ \ \ \ \ \ \ =2{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =2\ (0.3010)\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.6020\\\\ \text{(e)}\;\;{{\log }_{{10}}}4.5=\;{{\log }_{{10}}}\displaystyle \frac{9}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}\displaystyle \frac{{{{3}^{2}}}}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;2{{\log }_{{10}}}3-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;2\left( {0.4771} \right)-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;0.6532\\\\ \text{(f)}\;\;{{\log }_{{10}}}8\ ={{\log }_{{10}}}{{2}^{3}}\\ \ \ \ \ \ \ \ \ \ \ \ \ =3{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =3\left( {0.3010} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.9030\\\\ \text{(g)}\;\;{{\log }_{{10}}}18={{\log }_{{10}}}\left( {2\times {{3}^{2}}} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}2+2{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.3010+2\left( {0.4771} \right)\\\\ \text{(h)}\;\;{{\log }_{{10}}}5={{\log }_{{10}}}\displaystyle \frac{{10}}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}10-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =1-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.6990 \end{array}$

6.           Solve the following equations for $x$.

              $\begin{array}{lll} \text{(a)}\ \ \log _{a} \displaystyle\frac{18}{5}+\log _{a} \displaystyle\frac{10}{3}-\log _{a} \displaystyle\frac{6}{7}=\log _{a} x\\\\ \text{(b)}\ \ \log _{b} x=2-a+\log _{b}\left(\displaystyle\frac{a^{2} b^{a}}{b^{2}}\right)\\\\ \text{(c)}\ \ \log x^{3}-\log x^{2}=\log 5 x-\log 4 x\\\\ \text{(d)}\ \ \log _{10} x+\log _{10} 3=\log _{10} 6\\\\ \text{(e)}\ \ 8 \log x=\log a^{\frac{3}{2}}+\log 2-\displaystyle\frac{1}{2} \log a^{3}-\log \frac{2}{a^{4}}\\\\ \end{array}$

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$\begin{array}{l} \text{(a)}\;\;{{\log }_{a}}\displaystyle \frac{{18}}{5}+{{\log }_{a}}\displaystyle \frac{{10}}{3}-{{\log }_{a}}\displaystyle \frac{6}{7}={{\log }_{a}}x\\ \ \ \ \ {{\log }_{a}}\left( {\displaystyle \frac{{\displaystyle \frac{{18}}{5}\times \displaystyle \frac{{10}}{3}}}{{\displaystyle \frac{6}{7}}}} \right)={{\log }_{a}}x\\ \ \ \ \ x=\displaystyle \frac{{18}}{5}\times \displaystyle \frac{{10}}{3}\times \displaystyle \frac{7}{6}\\ \ \ \ \ x=14\\\\ \text{(b)}\;\;{{\log }_{b}}x=2-a+{{\log }_{b}}\left( {\displaystyle \frac{{{{a}^{2}}{{b}^{a}}}}{{{{b}^{2}}}}} \right)\\ \ \ \ \ {{\log }_{b}}x=2-a+{{\log }_{b}}{{a}^{2}}+{{\log }_{b}}{{b}^{a}}-{{\log }_{b}}{{b}^{2}}\\ \ \ \ \ {{\log }_{b}}x=2-a+{{\log }_{b}}{{a}^{2}}+a-2\\ \ \ \ \ {{\log }_{b}}x={{\log }_{b}}{{a}^{2}}\\ \ \ \ \ x={{a}^{2}}\\\\ \text{(c)}\;\;\log {{x}^{3}}-\log {{x}^{2}}=\log 5x-\log 4x\\ \ \ \ \ \log \displaystyle \frac{{{{x}^{3}}}}{{{{x}^{2}}}}=\log \displaystyle \frac{{5x}}{{4x}}\\ \ \ \ \ \log x=\log \displaystyle \frac{5}{4}\\ \ \ \ \ x=\displaystyle \frac{5}{4}\\\\ \text{(d)}\;\;{{\log }_{{10}}}x+{{\log }_{{10}}}3={{\log }_{{10}}}6\\ \ \ \ \ \ {{\log }_{{10}}}3x={{\log }_{{10}}}6\\ \ \ \ \ \ 3x=6\\ \ \ \ \ \ x=2\\ \text{(e)}\;\;8\log x=\log {{a}^{{\frac{3}{2}}}}+\log 2- \frac{1}{2}\log {{a}^{3}}-\log \displaystyle \frac{2}{{{{a}^{4}}}}\\ \ \ \ \ \log {{x}^{8}}=\log {{a}^{{\frac{3}{2}}}}+\log 2-\log {{a}^{{ \frac{3}{2}}}}-\left( {\log 2-\log {{a}^{4}}} \right)\\ \ \ \ \ \log {{x}^{8}}=\log {{a}^{4}}\\ \ \ \ \ {{x}^{8}}={{a}^{4}}\\ \ \ \ \ x=\sqrt{a}\end{array}$

7.           Given that $\log_{10} 5 = 0.6990$ and $\log_{10}x = 0.2330$. What is the value of $x$?

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$\begin{array}{l} \ \ \;\;{{\log }_{{10}}}5=0.6990\\\\ \ \ \ \ {{\log }_{{10}}}x=0.2330\\\\ \ \ \ \ \therefore \ \ 3{{\log }_{{10}}}x=0.6990\\\\ \ \ \ \ \therefore \ \ 3{{\log }_{{10}}}x={{\log }_{{10}}}5\\\\ \ \ \ \ \therefore \ \ {{\log }_{{10}}}x=\displaystyle\frac{1}{3}{{\log }_{{10}}}5\\\\ \ \ \ \ \therefore \ \ {{\log }_{{10}}}x={{\log }_{{10}}}{{5}^{{\frac{1}{3}}}}\\\\ \ \ \ \ \therefore \ \ x={{5}^{{\frac{1}{3}}}}=\sqrt[3]{5} \end{array}$

8.           Show that if $\log _{e} I=-\displaystyle\frac{R}{L} t+\log _{e} I_{0}$ then $I=I_{0} e^{-\frac{R t}{L}}$

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$\begin{array}{l} {{\log }_{e}}I=-\displaystyle \frac{R}{L}t+{{\log }_{e}}{{I}_{0}}\\\\ {{\log }_{e}}I-{{\log }_{e}}{{I}_{0}}=-\displaystyle \frac{{Rt}}{L}\\\\ {{\log }_{e}}\displaystyle \frac{I}{{{{I}_{0}}}}=-\displaystyle \frac{{Rt}}{L}\\\\ \displaystyle \frac{I}{{{{I}_{0}}}}={{e}^{{-\frac{{Rt}}{L}}}}\\\\ \therefore \ I={{I}_{0}}{{e}^{{-\frac{{Rt}}{L}}}} \end{array}$

9.           Show that if $\log _{b} y=\displaystyle\frac{1}{2} \log _{b} x+c$ then $y=b^{c} \sqrt{x}$

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$\begin{array}{l}{{\log }_{b}}y=\displaystyle \frac{1}{2}{{\log }_{b}}x+c\\\\ {{\log }_{b}}y-\displaystyle \frac{1}{2}{{\log }_{b}}x=c\\\\ {{\log }_{b}}y-{{\log }_{b}}{{x}^{{\frac{1}{2}}}}=c\\\\ {{\log }_{b}}\displaystyle \frac{y}{{\sqrt{x}}}=c\\\\ \displaystyle \frac{y}{{\sqrt{x}}}={{b}^{c}}\\\\y=\ {{b}^{c}}\sqrt{x} \end{array}$

10.           Show that

              $\begin{array}{l} \text{(a)}\ \ \displaystyle\frac{1}{4} \log _{10} 8+\frac{1}{4} \log _{10} 2=\log _{10} 2\\\\ \text{(b)}\ \ 4 \log _{10} 3-2 \log _{10} 3+1=\log _{10} 90 \end{array}$

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$\begin{array}{l}\text{(a)}\;\;\ \ \ \displaystyle \frac{1}{4}{{\log }_{{10}}}8+\displaystyle \frac{1}{4}{{\log }_{{10}}}2\\\ \ \ \ =\displaystyle \frac{1}{4}\left( {{{{\log }}_{{10}}}8+{{{\log }}_{{10}}}2} \right)\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}\left( {8\times 2} \right)\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}16\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}{{2}^{4}}\\\ \ \ \ =\displaystyle \frac{1}{4}\times 4{{\log }_{{10}}}2\\\ \ \ \ ={{\log }_{{10}}}2\\\\\text{(b)}\;\;\ \ 4{{\log }_{{10}}}3-2{{\log }_{{10}}}3+1\\\ \ \ \ =2{{\log }_{{10}}}3+{{\log }_{{10}}}10\\\ \ \ \ ={{\log }_{{10}}}{{3}^{2}}+{{\log }_{{10}}}10\\\ \ \ \ ={{\log }_{{10}}}\left( {{{3}^{2}}\times 10} \right)\\\ \ \ \ ={{\log }_{{10}}}90\end{array}$

11.           Show that

              $\begin{array}{l} \text{(a)}\ \ a^{2 \log _{a} 3}+b^{3 \log _{b} 2}=17\\\\ \text{(b)}\ \ 3 \log _{6} 1296=2 \log _{4} 4096 \end{array}$

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$\begin{array}{l}\text{(a)}\;\;\ \ {{a}^{{2{{{\log }}_{a}}3}}}+{{b}^{{3{{{\log }}_{b}}2}}}\\\ \ \ \ ={{a}^{{{{{\log }}_{a}}{{3}^{2}}}}}+{{b}^{{{{{\log }}_{b}}{{2}^{3}}}}}\\\ \ \ \ ={{3}^{2}}+{{2}^{3}}\\\ \ \ \ =17\\\\\text{(b)}\;\ \ 3{{\log }_{6}}1296=3{{\log }_{6}}{{6}^{4}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3\times 4{{\log }_{6}}6\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =12\\\ \ \ \ \ \ 2{{\log }_{4}}4096=2{{\log }_{4}}{{4}^{6}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\times 6{{\log }_{4}}4\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =12\\\ \ \ \therefore \ \ 3{{\log }_{6}}1296=2{{\log }_{4}}4096\end{array}$

12.           Given that $\log _{10} 12=1.0792$ and $\log _{10} 24=1.3802,$ deduce the values of $\log _{10} 2$ and $\log _{10} 6$.

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$\begin{array}{l}{{\log }_{{10}}}12=1.0792\\\\{{\log }_{{10}}}24=1.3802\\\\{{\log }_{{10}}}2={{\log }_{{10}}}\displaystyle \frac{{24}}{{12}}\\\ \ \ \ \ \ \ ={{\log }_{{10}}}24-{{\log }_{{10}}}12\\\ \ \ \ \ \ \ =1.3802-1.0792\\\ \ \ \ \ \ \ =0.3010\\\\{{\log }_{{10}}}6={{\log }_{{10}}}\displaystyle \frac{{12}}{2}\\\ \ \ \ \ \ \ ={{\log }_{{10}}}12-{{\log }_{{10}}}2\\\ \ \ \ \ \ \ =1.0792-0.3010\\\ \ \ \ \ \ \ =0.7782\end{array}$

13.           If $\log _{x} a=5$ and $\log _{x} 3 a=9$, find the values of $a$ and $x$.

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$\begin{array}{l}{{\log }_{x}}a=5\\\\\therefore \ \ a={{x}^{5}}\\\\{{\log }_{x}}3a=9\\\\\therefore \ \ 3a={{x}^{9}}\\\\\therefore \ \ 3{{x}^{5}}={{x}^{9}}\\\\\therefore \ \ {{x}^{4}}=3\\\\\therefore \ \ x={{3}^{{\frac{1}{4}}}}=\sqrt[4]{3}\\\\\therefore \ \ a={{\left( {{{3}^{{\frac{1}{4}}}}} \right)}^{5}}={{3}^{{\frac{5}{4}}}}=3\sqrt[4]{3}\end{array}$

14.           $\text { (a) }$ If $\log _{10} 2=a,$ find $\log _{10} 8+\log _{10} 25$ in terms of $a$.

   $\text { (b) }$ If $a=10^{x}$ and $b=10^{y},$ express $\log _{10}\left(a^{4} b^{3}\right)$ in terms of $x$ and $y$.

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$\begin{array}{l}(\text{a})\ {{\log }_{{10}}}2=a\ (\text{given})\\\\\ \ \ {{\log }_{{10}}}8+{{\log }_{{10}}}25\\\\={{\log }_{{10}}}8+{{\log }_{{10}}}\frac{{100}}{4}\\\\={{\log }_{{10}}}8+{{\log }_{{10}}}100-{{\log }_{{10}}}4\\\\={{\log }_{{10}}}{{2}^{3}}+{{\log }_{{10}}}{{10}^{2}}-{{\log }_{{10}}}{{2}^{2}}\\\\=3{{\log }_{{10}}}2+2{{\log }_{{10}}}10-2{{\log }_{{10}}}2\\\\={{\log }_{{10}}}2+2\\\\=a+2\\\\(\text{b})\ \ \left. \begin{array}{l}a={{10}^{x}}\\b={{10}^{y}}\end{array} \right\}(\text{given})\\\\\ \ \ \ \ {{\log }_{{10}}}\left( {{{a}^{4}}{{b}^{3}}} \right)={{\log }_{{10}}}\left( {{{{\left( {{{{10}}^{x}}} \right)}}^{4}}{{{\left( {{{{10}}^{y}}} \right)}}^{3}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}\left( {\left( {{{{10}}^{{4x}}}} \right)\left( {{{{10}}^{3}}^{y}} \right)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}{{10}^{{4x+3y}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4x+3y\end{array}$

15.           $\text { (a) }$ If $\log _{2}(4 x-4)=2$, find the value of $\log _{4} x$.

   $\text { (b) }$ Prove that if $\displaystyle\frac{1}{2} \log _{3} M+3 \log _{3} N=1$ then $M N^{6}=9$.

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$\begin{array}{l} (\text{a})\ {{\log }_{2}}(4x-4)=2\\\\ \ \ \ 4x-4={{2}^{2}}\\\\ \ \ \ 4x=8\\\\ \ \ \ x=2\\\\ \ \ \ x={{4}^{{\frac{1}{2}}}}\\\\ \ \ \ {{\log }_{4}}x=\displaystyle\frac{1}{2}\\\\\\ (\text{b})\ \ \displaystyle\frac{1}{2}{{\log }_{3}}M+3{{\log }_{3}}N=1\\\\ \ \ \ \ \ {{\log }_{3}}{{M}^{{\frac{1}{2}}}}+{{\log }_{3}}{{N}^{3}}=1\\\\ \ \ \ \ \ {{\log }_{3}}\left( {{{M}^{{\frac{1}{2}}}}{{N}^{3}}} \right)=1\\\\ \ \ \ \ \ {{M}^{{\frac{1}{2}}}}{{N}^{3}}=3\\\\ \ \ \ \ \ \text{Squaring both sides}\text{.}\\\\ \ \ \ \ \ M{{N}^{6}}=9 \end{array}$

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