Exponential Equations : Exercise (2.5) - Solutions

            Solve the following equations.

$\begin{array}{l} 1.\quad 3^{{2 x-3}}=27^{{2 x}} \\\\ 2.\quad 5^{{x^{2}-9}}=1 \\\\ 3.\quad 5^{{x+1}}=\displaystyle \frac{1}{625}\\\\ 4.\quad \left(\displaystyle \frac{1}{2}\right)^{x}=64 \\\\ 5.\quad 2^{{3 x}} \cdot 4^{{x+1}}=128 \\\\ 6.\quad 3^{{x+1}} \cdot 9^{2-x}=\displaystyle \frac{1}{27} \\\\ 7.\quad \displaystyle \frac{27^{{2 x}}}{3^{{5-x}}}=\displaystyle \frac{3^{{2 x+1}}}{9^{{x+3}}}\\\\ 8.\quad 8^{{x-1}}=\left(\displaystyle \frac{1}{32}\right)^{x+1}\\\\ 9.\quad 10^{{-x}}=0.000001 \\\\ 10.\quad 4^{{x}}+4^{{x+1}}=20 \\\\ 11.\quad 4 \cdot 2^{{2 x}}+3 \cdot 2^{{x}}-1=0 \end{array}$

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$\begin{array}{l}1.\quad\quad {{3}^{{2x-3}}}={{27}^{{2x}}}\\\ \ \ \ {{3}^{{2x-3}}}={{({{3}^{3}})}^{{2x}}}\\\ \ \ \ {{3}^{{2x-3}}}={{3}^{{6x}}}\\\ \therefore \ \ 2x-3=6x\\\ \therefore \ \ 8x=3\\\ \therefore \ \ x=\displaystyle \frac{3}{8}\\\\2.\quad\quad {{5}^{{{{x}^{2}}-9}}}=1\\\ \ \ \ {{5}^{{{{x}^{2}}-9}}}={{5}^{0}}\\\ \therefore \ \ {{x}^{2}}-9=0\\\ \ \ \ {{x}^{2}}=9\\\ \therefore \ \ x=\pm 3\\\\3.\quad\quad {{5}^{{x+1}}}=\displaystyle \frac{1}{{625}}\\\ \ \ \ {{5}^{{x+1}}}=\displaystyle \frac{1}{{{{5}^{4}}}}\\\ \ \ \ {{5}^{{x+1}}}={{5}^{{-4}}}\\\ \therefore \ \ x+1=-4\\\ \therefore \ \ x=-5\\\\4.\quad\quad {{\left( {\displaystyle \frac{1}{2}} \right)}^{x}}=64\\\ \ \ \ {{2}^{{-x}}}={{2}^{6}}\\\ \therefore \ \ -x=6\\\ \therefore \ \ x=6\\\\5.\quad\quad {{2}^{{3x}}}\cdot {{4}^{{x+1}}}=128\\\ \ \ \ {{2}^{{3x}}}\cdot {{\left( {{{2}^{2}}} \right)}^{{x+1}}}={{2}^{7}}\\\ \ \ \ {{2}^{{5x+2}}}={{2}^{7}}\\\ \therefore \ \ 5x+2=7\\\ \therefore \ \ 5x=5\\\ \therefore \ \ x=1\\\\6.\quad\quad {{3}^{{x+1}}}\cdot {{9}^{{2-x}}}=\displaystyle \frac{1}{{27}}\\\ \ \ \ {{3}^{{x+1}}}\cdot {{\left( {{{3}^{2}}} \right)}^{{2-x}}}=\displaystyle \frac{1}{{{{3}^{3}}}}\\\ \ \ \ {{3}^{{5-x}}}={{3}^{{-3}}}\\\ \therefore \ \ 5-x=-3\\\ \therefore \ \ x=8\\\ \therefore \ \ x=1\\\\7.\quad\quad \displaystyle \frac{{{{{27}}^{{2x}}}}}{{{{3}^{{5-x}}}}}=\displaystyle \frac{{{{3}^{{2x+1}}}}}{{{{9}^{{x+3}}}}}\\\ \ \ \ \displaystyle \frac{{{{{\left( {{{3}^{3}}} \right)}}^{{2x}}}}}{{{{3}^{{5-x}}}}}=\displaystyle \frac{{{{3}^{{2x+1}}}}}{{{{{\left( {{{3}^{2}}} \right)}}^{{x+3}}}}}\\\ \ \ \ {{3}^{{7x-5}}}={{3}^{{-5}}}\\\ \therefore \ \ 7x-5=-5\\\ \therefore \ \ x=0\\\ \therefore \ \ x=1\\\\8.\quad\quad {{8}^{{x-1}}}={{\left( {\displaystyle \frac{1}{{32}}} \right)}^{{x+1}}}\\\ \ \ \ {{\left( {{{2}^{3}}} \right)}^{{x-1}}}={{\left( {\displaystyle \frac{1}{{{{2}^{5}}}}} \right)}^{{x+1}}}\\\ \ \ \ {{2}^{{3x-3}}}={{2}^{{-5x-5}}}\\\ \therefore \ \ 3x-3=-5x-5\\\ \therefore \ \ 8x=8\\\ \therefore \ \ x=1\\\\9.\quad\quad {{10}^{{-x}}}=0.000001\\\ \ \ \ {{10}^{{-x}}}=\displaystyle \frac{1}{{1000000}}\\\ \ \ \ {{10}^{{-x}}}=\displaystyle \frac{1}{{{{{10}}^{6}}}}\\\ \ \ \ {{10}^{{-x}}}={{10}^{{-6}}}\\\ \therefore \ \ -x=-6\\\ \therefore \ \ x=6\\\\10.\;\ {{10}^{{-x}}}=0.000001\\\ \ \ \ {{10}^{{-x}}}=\displaystyle \frac{1}{{1000000}}\\\ \ \ \ {{10}^{{-x}}}=\displaystyle \frac{1}{{{{{10}}^{6}}}}\\\ \ \ \ {{10}^{{-x}}}={{10}^{{-6}}}\\\ \therefore \ \ -x=-6\\\ \therefore \ \ x=6\\\\11.\;\ 4\cdot {{2}^{{2x}}}+3\cdot {{2}^{x}}-1=0\\\ \ \ \ 4\cdot {{\left( {{{2}^{x}}} \right)}^{2}}+3\cdot {{2}^{x}}-1=0\\\ \ \ \ \text{Let}\ {{2}^{x}}=a,\ \text{then we have}\\\ \therefore \ \ 4{{a}^{2}}+3a-1=0\\\ \therefore \ \ (4a-1)(a+1)=0\\\ \therefore \ \ a=\displaystyle \frac{1}{4}\ \text{or}\ a=-1\\\ \therefore \ \ {{2}^{x}}=\displaystyle \frac{1}{4}\ \text{or}\ {{2}^{x}}=-1\ (\text{impossible})\\\ \ \ \ \text{Since}\ {{2}^{x}}>0,\ {{2}^{x}}=-1\ \text{is impossible}\text{.}\\\ \therefore \ \ {{2}^{x}}=\displaystyle \frac{1}{4}=\displaystyle \frac{1}{{{{2}^{2}}}}={{2}^{{-2}}}\\\therefore \ \ x=-2\end{array}$