# 6^x+2=4^x

Question
Logarithms
$$\displaystyle{6}^{{x}}+{2}={4}^{{x}}$$

2020-11-15

Take the natural logarithm of both sides:
$$\displaystyle{{\ln{{6}}}^{{x}}+}{2}={{\ln{{4}}}^{{x}}}$$
Apply Power Property of logarithms: $$\displaystyle{\log{{b}}}{x}^{{a}}={a}{\log{{b}}}{x}$$
$$\displaystyle{\left({x}+{2}\right)}{\ln{{6}}}={x}{\ln{{4}}}$$
$$\displaystyle{x}{\ln{{6}}}+{2}{\ln{{6}}}={x}{\ln{{4}}}$$
Subtract $$x\ln6$$ from both sides:
$$\displaystyle{2}{\ln{{6}}}={x}{\ln{{4}}}−{x}{\ln{{6}}}$$
Factor out x:
$$\displaystyle{2}{\ln{{6}}}={\left({\ln{{4}}}−{\ln{{6}}}\right)}{x}$$
Divide both sides by $$\displaystyle{\ln{{4}}}−{\ln{{6}}}{\ln{{4}}}−{\ln{{6}}}:$$
$$\displaystyle{2}\frac{{\ln{{6}}}}{{{\ln{{4}}}−{\ln{{6}}}}}={x}$$
or
$$\displaystyle{x}={2}\frac{{\ln{{6}}}}{{{\ln{{4}}}−{\ln{{6}}}}}≈−{8.84}$$

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