# 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.
$$\displaystyle{10}^{{{6}-{3}{x}}}={18}$$

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Given that:
The equation $$\displaystyle{10}^{{{6}-{3}{x}}}-{18}$$.
By using,
Natural Logarithm rule:
Exponent rule,
$$\displaystyle{\ln{{\left({x}^{{y}}\right)}}}={y}{\ln{{\left({x}\right)}}}$$
To solve the given equation.
By using the above formula,
Taking natural log on both side,
$$\displaystyle{\ln{{\left({10}^{{{6}-{3}{x}}}\right)}}}={\ln{{\left({18}\right)}}}$$
$$\displaystyle{\left({6}-{3}{x}\right)}{\ln{{\left({10}\right)}}}={\ln{{\left({18}\right)}}}$$
Divide on both side by $$\displaystyle{\ln{{\left({10}\right)}}}$$
To get,
$$\displaystyle{6}-{3}{x}={\frac{{{\ln{{\left({18}\right)}}}}}{{{\ln{{\left({10}\right)}}}}}}$$
$$\displaystyle-{3}{x}={\frac{{{\ln{{\left({18}\right)}}}}}{{{\ln{{\left({10}\right)}}}}}}-{6}$$
$$\displaystyle{x}={\frac{{{1}}}{{-{3}}}}{\frac{{{\ln{{\left({18}\right)}}}}}{{{\ln{{\left({10}\right)}}}}}}+{3}$$
$$\displaystyle{x}={\frac{{{1}}}{{-{3}}}}{\left({1.2553}\right)}+{3}$$
$$\displaystyle={\frac{{{1.2553}}}{{-{3}}}}+{3}$$
$$\displaystyle=-{0.4184}+{3}$$
$$\displaystyle={2.5816}$$
To get,
$$\displaystyle{x}={2.5816}$$
Therefore, $$\displaystyle{x}={2.5816}$$