Logarithms : Exercise (3.4) - Solutions

1.          If $\log _{a} b+\log _{b} a^{2}=3,$ find $b$ in terms of $a$.

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$\begin{array}{l}{{\log }_{a}}\,b+{{\log }_{b}}\,{{a}^{2}}=3\\\\{{\log }_{a}}\,b+2{{\log }_{b}}a=3\\\\{{\log }_{a}}\,b+\displaystyle\frac{2}{{{{{\log }}_{a}}\,b}}=3\\\\\text{Let}\ {{\log }_{a}}\,b=u,\ \text{then}\\\\u+\displaystyle\frac{2}{u}=3\\\\\therefore \ \ {{u}^{2}}+2=3u\\\\\therefore \ \ {{u}^{2}}-3u+2=0\\\\\therefore \ \ (u-1)(u-2)=0\\\\\therefore \ \ {{\log }_{a}}\,b=1\ \text{or}\ {{\log }_{a}}\,b=2\\\\\therefore \ \ \,b=a\ \text{or}\ \,b={{a}^{2}}\end{array}$

2.          Show that

(a) $\log _{4} x=2 \log _{16} x$.

(b) $\log _{b} x=3 \log _{b^{3}} x \quad$.

(c) $\log _{2} x=\left(1+\log _{2} 3\right) \log _{6} x$.

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$\begin{array}{l}(\text{a})\ \ \ \text{LHS}={{\log }_{4}}x\\\\\ \ \ \ \ \ \text{RHS}=2{{\log }_{{16}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{\log }_{{{{4}^{2}}}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{2}{2}{{\log }_{4}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{4}}x\\\\\ \ \therefore \ \ {{\log }_{4}}x=2{{\log }_{{16}}}x\\\\\\(\text{b})\ \ \text{LHS}={{\log }_{b}}x\\\\\ \ \ \ \ \text{RHS}=3{{\log }_{{{{b}^{3}}}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{3}{3}{{\log }_{b}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}x\\\\\ \ \therefore \ \ {{\log }_{b}}x=3{{\log }_{{{{b}^{3}}}}}x\\\\\\(\text{c})\ \ \text{LHS}={{\log }_{2}}x\\\\\ \ \ \ \ \text{RHS}=\left( {1+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {{{{\log }}_{2}}2+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{2}}6\cdot {{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{{{{{\log }}_{6}}2}}\cdot {{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{{{{{\log }}_{6}}x}}{{{{{\log }}_{6}}2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{2}}x\\\\\ \ \therefore \ \ {{\log }_{2}}x=\left( {1+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\end{array}$

3.          If $a=\log _{b} c, b=\log _{c} a$ and $c=\log _{a} b,$ prove that $a b c=1$.

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$ \begin{array}{l}\ \ \ a={{\log }_{b}}c,\ b={{\log }_{c}}a,\ c={{\log }_{a}}b\\\\\therefore \ \ abc={{\log }_{b}}c\cdot \ {{\log }_{c}}a\cdot {{\log }_{a}}b\\\\\therefore \ \ abc=\displaystyle\frac{{\log c}}{{\log b}}\cdot \ \displaystyle\frac{{\log a}}{{\log c}}\cdot \displaystyle\frac{{\log b}}{{\log a}}\ \left[ {\because \ \text{by L8}} \right]\\\\\therefore \ \ abc=1\end{array}$

4.          Show that

(a) $\left(\log _{10} 4-\log _{10} 2\right) \log _{2} 10=1$.

(b) $2 \log _{2} 3\left(\log _{9} 2+\log _{9} 4\right)=3$.

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$\begin{array}{l}\text{(a)}\ \ \ \left( {{{{\log }}_{{10}}}4-{{{\log }}_{{10}}}2} \right)\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}\displaystyle\frac{4}{2}\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}2\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}2\cdot \displaystyle\frac{1}{{{{{\log }}_{{10}}}2}}\\\\\ \ \ =\ 1\\\\\\\text{(b)}\ \ \ 2\cdot {{\log }_{2}}3\cdot \left( {{{{\log }}_{9}}2+{{{\log }}_{9}}4} \right)\\\\\ \ \ ={{\log }_{2}}{{3}^{2}}\cdot {{\log }_{9}}(2\times 4)\\\\\ \ \ ={{\log }_{2}}9\cdot {{\log }_{9}}8\\\\\ \ \ ={{\log }_{2}}9\cdot {{\log }_{9}}{{2}^{3}}\\\\\ \ \ ={{\log }_{2}}9\cdot 3{{\log }_{9}}2\\\\\ \ \ =3\cdot \displaystyle\frac{1}{{{{{\log }}_{9}}2}}\cdot {{\log }_{9}}2\\\\\ \ \ =3\end{array}$

5.          Compute

(a) $3^{\log _{2} 5}-5^{\log _{2} 3}$

(b) $4^{\log _{2} 3}$

(c) $2^{\log _{2 \sqrt{2}} 27}$

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$\begin{array}{l}\text{(a)}\ \ \ {{3}^{{{{{\log }}_{2}}\,5}}}-{{5}^{{{{{\log }}_{2}}\,3}}}\\\\\ \ \ \ ={{3}^{{{{{\log }}_{2}}\,5}}}-{{3}^{{{{{\log }}_{2}}\,5}}}\\\\\ \ \ \ =0\\\\\text{(b)}\ \ \ {{4}^{{{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{2}}} \right)}^{{{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{{{{{\log }}_{2}}3}}}} \right)}^{2}}\\\\\ \ \ \ ={{3}^{2}}\\\\\ \ \ \ =9\\\\\\\text{(c)}\ \ \ {{2}^{{{{{\log }}_{{2\sqrt{2}}}}27}}}\\\\\ \ \ \ ={{2}^{{{{{\log }}_{{{{2}^{{3/2}}}}}}{{3}^{3}}}}}\\\\\ \ \ \ ={{2}^{{\frac{2}{3}\times 3{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{2}^{{2{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{{{{{\log }}_{2}}3}}}} \right)}^{2}}\\\\\ \ \ \ ={{3}^{2}}\\\\\ \ \ \ =9\end{array}$

6.          Using properties of logarithm, solve the following equations.

(a) $ {{\log }_{x}}10+{{\log }_{{{{x}^{2}}}}}10=2$

(b) $25^{\log _{10} x}=5+4 x^{\log _{10} 5}$

(c) $9^{\log _{3}\left(\log _{2} x\right)}=\log _{2} x-\left(\log _{2} x\right)^{2}+1$

(d) $\log _{2} x+\log _{4} x+\log _{16} x=7$

(e) $\log _{7} 2+\log _{49} x=\log _{\frac{1}{7}} \sqrt{3}$

(f) $\log _{3} x-\log _{\frac{1}{3}} x^{2}=6$

(g) $ {{\log }_{x}}(9{{x}^{2}})\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4$

No (6) သည် ပြဌာန်းစာအုပ်တွင် မပါဝင်ပါ။ Change of Base နှင့် သက်ဆိုင်သော ဉာဏ်စမ်းမေးခွန်း များကို ထပ်ဆင့်လေ့ကျင့်နိုင်ရန် ပေါင်းထည့်ပေးထားခြင်း ဖြစ်ပါသည်။

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$ \begin{array}{l} \text{(a)}\ \ {{\log }_{x}}10+{{\log }_{{{{x}^{2}}}}}10=2\\\\ \ \ \ \ \ {{\log }_{x}}10+\displaystyle\frac{1}{2}{{\log }_{x}}10=2\\\\ \ \ \ \ \ \displaystyle\frac{3}{2}{{\log }_{x}}10=2\\\\ \ \ \ \ \ {{\log }_{x}}10=\displaystyle\frac{4}{3}\\\\ \ \ \ \ \ {{x}^{{\frac{4}{3}}}}=10\\\\ \ \ \ \ \ x={{10}^{{\frac{3}{4}}}}\\\\\\ \text{(b)}\ \ {{25}^{{{{{\log }}_{{10}}}x}}}=5+4{{x}^{{{{{\log }}_{{10}}}5}}}\\\\ \ \ \ \ \ {{\left( {{{5}^{2}}} \right)}^{{{{{\log }}_{{10}}}x}}}=5+4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}\\\\ \ \ \ \ \ {{5}^{{2\cdot }}}^{{{{{\log }}_{{10}}}x}}=5+4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}\\\\ \ \ \ \ \ {{\left( {{{5}^{{{{{\log }}_{{10}}}x}}}} \right)}^{2}}-4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}-5=0\\\\ \ \ \ \ \ \text{Let}\ {{5}^{{{{{\log }}_{{10}}}x}}}=u,\ \text{then}\\\\ \ \ \ \ \ {{u}^{2}}-4u-5=0\\\\ \ \ \ \ \ (u-5)(u+1)=0\\\\ \ \therefore \ \ \ u=5\ \text{or}\ u=-1\\\\ \ \therefore \ \ \ {{5}^{{{{{\log }}_{{10}}}x}}}=5\ \text{or}\ {{5}^{{{{{\log }}_{{10}}}x}}}=-1\\\\ \ \ \ \ \ \text{Since}\ {{5}^{{{{{\log }}_{{10}}}x}}}>0,\ {{5}^{{{{{\log }}_{{10}}}x}}}=-1\ \text{is}\ \text{impossible}\text{.}\\\\ \ \therefore \ \ \ {{5}^{{{{{\log }}_{{10}}}x}}}=5\\\\ \ \therefore \ \ \ {{\log }_{{10}}}x=1\\\\ \ \therefore \ \ \ x=10\\\\\\ \text{(c)}\ \ {{9}^{{{{{\log }}_{3}}\left( {{{{\log }}_{2}}x} \right)}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{{{{{\log }}_{3}}9}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{{{{{\log }}_{3}}{{3}^{2}}}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{2}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\ \ \ \ \ \ 2{{\left( {{{{\log }}_{2}}x} \right)}^{2}}-{{\log }_{2}}x-1=0\\\\ \ \ \ \ \ \left( {2\cdot {{{\log }}_{2}}x+1} \right)\left( {{{{\log }}_{2}}x-1} \right)=0\\\\ \ \ \ \ \ {{\log }_{2}}x=-\displaystyle\frac{1}{2}\ \text{or}\ {{\log }_{2}}x=1\\\\ \ \therefore \ \ \ x={{2}^{{-\frac{1}{2}}}}\ \text{or}\ x=2\\\\ \ \therefore \ \ \ x=\displaystyle\frac{1}{{\sqrt{2}}}\ \text{or}\ x=2\\\\\\ \text{(d)}\ \ {{\log }_{2}}x+{{\log }_{4}}x+{{\log }_{{16}}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x+{{\log }_{{{{2}^{2}}}}}x+{{\log }_{{{{2}^{4}}}}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x+\displaystyle\frac{1}{2}{{\log }_{2}}x+\displaystyle\frac{1}{4}{{\log }_{2}}x=7\\\\ \ \ \ \ \ \displaystyle\frac{7}{4}{{\log }_{2}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x=4\\\ \therefore \ \ \ x={{2}^{4}}=16\\\\\\ \text{(e)}\ \ {{\log }_{7}}2+{{\log }_{{49}}}x={{\log }_{{\displaystyle\frac{1}{7}}}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+{{\log }_{{{{7}^{2}}}}}x={{\log }_{{{{7}^{{-1}}}}}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+\displaystyle\frac{1}{2}{{\log }_{7}}x=-{{\log }_{7}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+{{\log }_{7}}{{x}^{{\frac{1}{2}}}}={{\log }_{7}}{{3}^{{-\frac{1}{2}}}}\\\\ \ \ \ \ \ {{\log }_{7}}2{{x}^{{\frac{1}{2}}}}={{\log }_{7}}\displaystyle\frac{1}{{{{3}^{{\frac{1}{2}}}}}}\\\\ \ \ \therefore \ \ 2{{x}^{{\frac{1}{2}}}}=\displaystyle\frac{1}{{{{3}^{{\frac{1}{2}}}}}}\\\\ \ \ \therefore \ \ 4x=\displaystyle\frac{1}{3}\\\ \ \therefore \ \ x=\displaystyle\frac{1}{{12}}\\\\\\ \text{(f)}\ \ {{\log }_{3}}x-{{\log }_{{\displaystyle\frac{1}{3}}}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}x-{{\log }_{{{{3}^{{-1}}}}}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}x+{{\log }_{3}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}{{x}^{3}}=6\\\\ \ \ \ \ \ {{x}^{3}}={{3}^{6}}\\\\ \ \therefore \ \ \ x={{3}^{2}}=9\\\\\\ \text{(g)}\ \ {{\log }_{x}}(9{{x}^{2}})\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4\\\\ \ \ \ \ \ \displaystyle\frac{{{{{\log }}_{3}}(9{{x}^{2}})}}{{{{{\log }}_{3}}x}}\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4\\\\ \ \ \ \ \ {{\log }_{3}}(9{{x}^{2}})\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ ({{\log }_{3}}9+{{\log }_{3}}{{x}^{2}})\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ ({{\log }_{3}}{{3}^{2}}+2{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ (2+2{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \therefore \ \ (1+{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=2\\\\ \ \therefore \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}+\left( {{{{\log }}_{3}}x} \right)-2=0\\\\ \ \therefore \ \ ({{\log }_{3}}x+2)({{\log }_{3}}x-1)=0\\\\ \ \therefore \ \ \ {{\log }_{3}}x=-2\ \text{or}\ {{\log }_{3}}x=1\\\\ \ \therefore \ \ \ x={{3}^{{-2}}}\ \text{or}\ x=3\\\\ \ \therefore \ \ \ x=\displaystyle\frac{1}{9}\ \text{or}\ x=3 \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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$\begin{array}{l} (\text{a})\ {{\log }_{2}}(4x-4)=2\\\\ \ \ \ 4x-4={{2}^{2}}\\\\ \ \ \ 4x=8\\\\ \ \ \ x=2\\\\ \ \ \ x={{4}^{{\frac{1}{2}}}}\\\\ \ \ \ {{\log }_{4}}x=\displaystyle\frac{1}{2}\\\\\\ (\text{b})\ \ \displaystyle\frac{1}{2}{{\log }_{3}}M+3{{\log }_{3}}N=1\\\\ \ \ \ \ \ {{\log }_{3}}{{M}^{{\frac{1}{2}}}}+{{\log }_{3}}{{N}^{3}}=1\\\\ \ \ \ \ \ {{\log }_{3}}\left( {{{M}^{{\frac{1}{2}}}}{{N}^{3}}} \right)=1\\\\ \ \ \ \ \ {{M}^{{\frac{1}{2}}}}{{N}^{3}}=3\\\\ \ \ \ \ \ \text{Squaring both sides}\text{.}\\\\ \ \ \ \ \ M{{N}^{6}}=9 \end{array}$

Logarithms : Exercise (3.2) Solutions

1.           Write the following equations in logarithmic form.

              (a)   $3^{4}=81$

              (b)   $9^{\frac{3}{2}}=27$

              (c)   $10^{-3}=0.001$

              (d)   $3^{-1}=\displaystyle\frac{1}{3}$

              (e)   $\left(\displaystyle\frac{1}{4}\right)^{-3}=64$

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$\begin{array}{l} \text{(a)}\ \ {{3}^{4}}=81\\ \ \ \ \ \ {{\log }_{3}}81=4\\\\ \text{(b)}\ \ {{9}^{{\frac{3}{2}}}}=27\\ \ \ \ \ \ {{\log }_{9}}27=\displaystyle\frac{3}{2}\\\\ \text{(c)}\ \ {{10}^{{-3}}}=0.001\\ \ \ \ \ \ {{\log }_{{10}}}0.001=-3\\\\ \text{(d)}\ \ {{3}^{{-1}}}=\displaystyle\frac{1}{3}\\ \ \ \ \ \ {{\log }_{3}}\displaystyle\frac{1}{3}=-1\\\\ \text{(e)}\ \ {{\left( {\displaystyle\frac{1}{4}} \right)}^{{-3}}}=64\\ \ \ \ \ \ {{\log }_{{\frac{1}{4}}}}64=-3 \end{array}$

2.           Write the following equations in exponential form.

             (a)   $\log _{10} 3=0.4771$

             (b)   $\log _{6} 0.001=-3.855$

             (c)   $\log _{144} 12=\displaystyle\frac{1}{2}$

             (d)   $-5=\log _{3}\displaystyle \frac{1}{243}$

             (e)   $\log _{x} .7=y^{2},$ where, $0<x<1$

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$\begin{array}{l} \text{(a)}\ \ {{\log }_{{10}}}3=0.4771\\ \ \ \ \ \ {{3}^{{0.4771}}}=10\\\\ \text{(b)}\ \ {{\log }_{6}}0.001=-3.855\\ \ \ \ \ \ {{6}^{{-3.855}}}=0.001\\\\ \text{(c)}\ \ {{\log }_{{144}}}12=\displaystyle \frac{1}{2}\\ \ \ \ \ \ {{144}^{{\frac{1}{2}}}}=12\\\\ \text{(d)}\ \ -5={{\log }_{3}}\frac{1}{{243}}\\ \ \ \ \ \ \displaystyle \frac{1}{{243}}={{3}^{{-5}}}\\\\ \text{(e)}\ \ {{\log }_{x}}.7={{y}^{2}},\ \ \text{where}\ 0<x<1\\ \ \ \ \ {{x}^{{{{y}^{2}}}}}=.7 \end{array}$

3.           Solve the following equations

             (a)   $\log _{7} 49=x$

             (b)   $\log _{x} 10=1$

             (c)   $\log _{\sqrt{3}} x=2$

             (d)   $x^{\log _{x} x}=5$

             (e)   $\log _{0.2} 5=x$

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$\begin{array}{l} \text{(a)}\ \ {{\log }_{7}}49=x\\ \ \ \ \ \ {{7}^{x}}=49\\ \ \ \ \ \ {{7}^{x}}={{7}^{2}}\\ \ \ \ \ \ x=2\\\\ \text{(b)}\ \ {{\log }_{x}}10=1\\ \ \ \ \ \ {{x}^{1}}=10\\ \ \ \ \ \ x=10\\\\ \text{(c)}\ \ {{\log }_{{\sqrt{3}}}}x=2\ \\ \ \ \ \ x={{\left( {\sqrt{3}} \right)}^{2}}\\ \ \ \ \ \ x=3\\\\ \text{(d)}\ \ {{x}^{{{{{\log }}_{x}}x}}}=5\\ \ \ \ \ \ x=5\\\\ \text{(e)}\ \ {{\log }_{{0.2}}}5=x\\ \ \ \ \ \ {{0.2}^{x}}=5\\ \ \ \ \ \ {{\left( {\displaystyle\frac{1}{5}} \right)}^{x}}=5\\ \ \ \ \ \ {{5}^{{-x}}}={{5}^{1}}\\\ \ \ \ \ x=-1\ \ \end{array}$

4.           Evaluate.

             (a)   $9^{\log _{9} 2}+3^{\log _{3} 8}$

             (b)   $\log _{4} 4^{5} \times \log _{10} 10^{2}$

             (c)   $7^{\log _{7} 9}+\log _{2}\left(\displaystyle\frac{1}{2}\right)$

             (d)   $\log _{\frac{1}{2}} \displaystyle\frac{1}{8}-4 \log _{10} 10$

             (e)   $10^{1-\log _{10} 3}$

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$\begin{array}{l} \text{(a)}\ \ \ \ {{9}^{{{{{\log }}_{9}}2}}}+{{3}^{{{{{\log }}_{3}}8}}}\\ \ \ \ \ \ =2+8\\\ \ \ \ \ =10\\\\ \text{(b)}\ \ \ \ {{\log }_{4}}{{4}^{5}}\times {{\log }_{{10}}}{{10}^{2}}\\ \ \ \ \ \ =5\times 2\\\ \ \ \ \ =10\\\\ \text{(c)}\ \ \ \ \ {{7}^{{{{{\log }}_{7}}9}}}+{{\log }_{2}}\left( {\displaystyle\frac{1}{2}} \right)\\ \ \ \ \ \ ={{7}^{{{{{\log }}_{7}}9}}}+{{\log }_{2}}{{2}^{{-1}}}\\ \ \ \ \ \ =9-1\\\ \ \ \ \ =8\\\\ \text{(d)}\ \ \ \ \ {{\log }_{{\frac{1}{2}}}}\displaystyle\frac{1}{8}-4\cdot {{\log }_{{10}}}10\\ \ \ \ \ \ ={{\log }_{{\frac{1}{2}}}}{{\left( {\displaystyle\frac{1}{2}} \right)}^{3}}-4\cdot {{\log }_{{10}}}10\\ \ \ \ \ \ =3-4\cdot (1)\\\ \ \ \ \ =-1\\\\ \text{(e)}\ \ \ \ \ {{10}^{{1-{{{\log }}_{{10}}}3}}}\\ \ \ \ \ \ =\displaystyle\frac{{10}}{{{{{10}}^{{{{{\log }}_{{10}}}3}}}}}\\ \ \ \ \ \ =\displaystyle\frac{{10}}{3} \end{array}$

5.           Find the value of $x$ in each of the following problems.

             (a)   $\log _{3}(2 x-5)=2,$ where $x>\displaystyle\frac{5}{2}$

             (b)   $\log _{77}\left(\log _{7} x\right)=0,$ where $x>0$

             (c)   $8+3^{x}=10,$ given that $\log _{3} 2=0.6309$

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$\begin{array}{l}\text{(a)}\ \ {{\log }_{3}}(2x-5)=2,\text{where}\ x>\displaystyle \frac{5}{2}\\\ \ \ \ \ 2x-5={{3}^{2}}\\\ \ \ \ \ 2x-5=9\\\ \ \ \ \ x=7\\\\\text{(b)}\ \ {{\log }_{{77}}}\left( {{{{\log }}_{7}}x} \right)=0,\ \ \text{where}\ x>0\\\ \ \ \ \ {{\log }_{7}}x={{77}^{0}}\\\ \ \ \ \ {{\log }_{7}}x=1\\\ \ \ \ \ x={{7}^{1}}=7\\\\\text{(c)}\ \ 8+{{3}^{x}}=10,\ \text{given that}\ {{\log }_{3}}2=0.6309\\\ \ \ \ \ {{3}^{x}}=2\\\ \ \ \ \ x={{\log }_{3}}2\\\ \ \ \ \ x=0.6309\end{array}$

Logarithms

 John Napier (inventor of logarithms)

Logarithm ကို အေျခခံက်က်ေလး ရွင္းျပေပးထားလို႔ နားလည္လြယ္မွာပါ။ ေလ့က်င့္ဖို႔ exercises ေတြလည္းပါေတာ့ အသံုး၀င္မယ္လို႔ ေမွ်ာ္လင့္ပါတယ္။