# Find the exact solution of the exponential equation in terms

Find the exact solution of the exponential equation in terms of logarithms.
Use a calculator to find an approximation to the solution rounded to six decimal places.
I keep getting a negative log answer and decimal...i don't think that is correct...can i get some help please.
$$\displaystyle{8}^{{{1}-{x}}}={9}$$

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Feas1981
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:−
$$\displaystyle\Rightarrow{\ln{{\left({8}\right)}}}−{x}{\ln{{\left({8}\right)}}}={\ln{{\left({9}\right)}}}$$
$$\displaystyle\Rightarrow{x}{\ln{{\left({8}\right)}}}={\ln{{\left({9}\right)}}}−{\ln{{\left({8}\right)}}}$$
$$\displaystyle\Rightarrow{x}={\frac{{{\ln{{\left({9}\right)}}}-{\ln{{\left({8}\right)}}}}}{{{\ln{{\left({9}\right)}}}}}}$$
$$\displaystyle\Rightarrow{x}={\frac{{{2.197224}-{2.079441}}}{{{2.079441}}}}$$
$$\displaystyle\Rightarrow{x}={\frac{{{0.117783}}}{{{2.079441}}}}$$
$$\displaystyle\Rightarrow{x}={0.056642}\ \text{ is the exact solution of exponential equation }\ {8}^{{{1}-{x}}}={9}\ \text{ upto 6 decimal places.}$$
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Ourst1977
Thank you very much for the solution, I have been looking for it for a long time and could not find