Our aim is to find the exact the exact solution of the exponential equation

\(\displaystyle{8}^{{{1}-{x}}}={9}-{\left({i}\right)}\) in terms of logarithms.

Taking the logarithmic of equation−(i) both sides we have:−

\(\displaystyle{\ln{{\left({8}^{{{1}-{x}}}\right)}}}={\ln{{\left({9}\right)}}}-{\left({i}{i}\right)}\)

Now, using Logarithmic Power Rule in L.H.S of equation (ii), we have:−

\(\displaystyle{{\log}_{{b}}{\left({x}^{{y}}\right)}}={y}{{\log}_{{b}}{\left({x}\right)}}\rightarrow{\left[\text{Logarithmic Power Rule}\right]}\)

\(\displaystyle\Rightarrow{\left({1}−{x}\right)}{\ln{{\left({8}\right)}}}={\ln{{\left({9}\right)}}}−{\left({i}{i}{i}\right)}\)

Applying Distributive Property in L.H.S. of equation (iii), we have:−

\(\displaystyle\Rightarrow{\ln{{\left({8}\right)}}}−{x}{\ln{{\left({8}\right)}}}={\ln{{\left({9}\right)}}}\)

\(\displaystyle\Rightarrow{x}{\ln{{\left({8}\right)}}}={\ln{{\left({9}\right)}}}−{\ln{{\left({8}\right)}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{{\ln{{\left({9}\right)}}}-{\ln{{\left({8}\right)}}}}}{{{\ln{{\left({9}\right)}}}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{{2.197224}-{2.079441}}}{{{2.079441}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{{0.117783}}}{{{2.079441}}}}\)

\(\displaystyle\Rightarrow{x}={0.056642}\ \text{ is the exact solution of exponential equation }\ {8}^{{{1}-{x}}}={9}\ \text{ upto 6 decimal places.}\)

\(\displaystyle{8}^{{{1}-{x}}}={9}-{\left({i}\right)}\) in terms of logarithms.

Taking the logarithmic of equation−(i) both sides we have:−

\(\displaystyle{\ln{{\left({8}^{{{1}-{x}}}\right)}}}={\ln{{\left({9}\right)}}}-{\left({i}{i}\right)}\)

Now, using Logarithmic Power Rule in L.H.S of equation (ii), we have:−

\(\displaystyle{{\log}_{{b}}{\left({x}^{{y}}\right)}}={y}{{\log}_{{b}}{\left({x}\right)}}\rightarrow{\left[\text{Logarithmic Power Rule}\right]}\)

\(\displaystyle\Rightarrow{\left({1}−{x}\right)}{\ln{{\left({8}\right)}}}={\ln{{\left({9}\right)}}}−{\left({i}{i}{i}\right)}\)

Applying Distributive Property in L.H.S. of equation (iii), we have:−

\(\displaystyle\Rightarrow{\ln{{\left({8}\right)}}}−{x}{\ln{{\left({8}\right)}}}={\ln{{\left({9}\right)}}}\)

\(\displaystyle\Rightarrow{x}{\ln{{\left({8}\right)}}}={\ln{{\left({9}\right)}}}−{\ln{{\left({8}\right)}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{{\ln{{\left({9}\right)}}}-{\ln{{\left({8}\right)}}}}}{{{\ln{{\left({9}\right)}}}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{{2.197224}-{2.079441}}}{{{2.079441}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{{0.117783}}}{{{2.079441}}}}\)

\(\displaystyle\Rightarrow{x}={0.056642}\ \text{ is the exact solution of exponential equation }\ {8}^{{{1}-{x}}}={9}\ \text{ upto 6 decimal places.}\)