# Solve for the following exponential equations. Use the natural logarithm in your

Solve for the following exponential equations. Use the natural logarithm in your answer(where applicable) for full credit. Use rules for exponents, factor and simplify.
$$\displaystyle{4}^{{{1}-{x}}}={3}^{{{2}{x}+{5}}}$$

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Aamina Herring
We have,
$$\displaystyle{4}^{{{1}-{x}}}={3}^{{{2}{x}+{5}}}$$
Taking logarithm on both of the side and solving further, we get result as
$$\displaystyle{\left({1}-{x}\right)}{\log{{4}}}={\left({2}{x}+{5}\right)}{\log{{3}}}$$
$$\displaystyle{\log{{4}}}-{\left({\log{{4}}}\right)}{x}={\left({2}{\log{{3}}}\right)}{x}+{5}{\log{{3}}}$$
Taking the term related x on one of the side,
$$\displaystyle{\left({\log{{4}}}-{5}{\log{{3}}}\right)}={\left({2}{\log{{3}}}+{\log{{4}}}\right)}{x}$$
$$\displaystyle{\log{{4}}}-{{\log{{3}}}^{{5}}=}{\left({{\log{{3}}}^{{2}}+}{\log{{4}}}\right)}{x}$$
$$\displaystyle{x}={\frac{{{\left({\log{{4}}}-{\log{{243}}}\right)}}}{{{\left({\log{{9}}}+{\log{{4}}}\right)}}}}$$
Using the logarithm property $$\displaystyle{\log{{m}}}-{\log{{n}}}={\log{{\left({\frac{{{m}}}{{{n}}}}\right)}}}$$ and $$\displaystyle{\log{{m}}}+{\log{{n}}}={\log{{\left({m}\cdot{n}\right)}}}$$ and solving further, we get the result as
$$\displaystyle{x}={\frac{{{\left({\log{{4}}}-{\log{{243}}}\right)}}}{{{\left({\log{{9}}}+{\log{{4}}}\right)}}}}$$
$$\displaystyle{x}={\frac{{{\log{{\left({\frac{{{4}}}{{{243}}}}\right)}}}}}{{{\log{{36}}}}}}$$
$$\displaystyle{x}={{\log}_{{{36}}}{\left({\frac{{{4}}}{{{243}}}}\right)}}$$
Hence, value of x will be $$\displaystyle{x}={{\log}_{{{36}}}{4243}}$$,