Question

# Find the exact solution if possible: \log_{3}(x-8)+\log_{3}x=2

Decimals
Exponential and Logarithmic Equations Solve the equation. Find the exact solution if possible. otherwise, use a calculator to approximate to two decimals.
$$\displaystyle{{\log}_{{{3}}}{\left({x}-{8}\right)}}+{{\log}_{{{3}}}{x}}={2}$$

2021-08-04
Step 1
For solving Exponential Equations:
1) Isolate the exponential expression on one side of the equation.
2) Take the logarithm of each side, then use the Laws of Logarithms to "bring down exponent".
3) Solve for the variable.
Step 2
Given,
$$\displaystyle{{\log}_{{{3}}}{\left({x}-{8}\right)}}+{{\log}_{{{3}}}{x}}={2}$$
By the first law of logarithms function,
$$\displaystyle{{\log}_{{{3}}}{\left({x}-{8}\right)}}{x}={2}$$
By the definition of logarithm function,
$$\displaystyle{{\log}_{{{a}}}{x}}={y}\Leftrightarrow\ {a}^{{{y}}}={x}$$
$$\displaystyle{3}^{{{2}}}={x}{\left({x}-{8}\right)}$$
$$\displaystyle{9}={x}{\left({x}-{8}\right)}$$
By the distributive property of multiplication,
$$\displaystyle{x}\cdot\ {x}-{x}\cdot{8}={9}$$
$$\displaystyle{x}^{{{2}}}-{8}{x}={9}$$
Add $$\displaystyle-{9}$$ on both sides,
$$\displaystyle{x}^{{{2}}}-{8}{x}-{9}={9}-{9}$$
$$\displaystyle{x}^{{{2}}}-{8}{x}-{9}={0}$$
Factor $$\displaystyle{x}^{{{2}}}-{8}{x}-{9}$$ as $$\displaystyle{\left({x}-{9}\right)}{\left({x}+{1}\right)}.$$
$$\displaystyle{\left({x}-{9}\right)}{\left({x}+{1}\right)}={0}$$
By the zero-factor property,
$$\displaystyle{x}-{9}={0}$$ and $$\displaystyle{x}+{1}={0}$$
Taking,
$$\displaystyle{x}-{9}={0}$$
$$\displaystyle{x}-{9}+{9}={0}+{9}$$
$$\displaystyle{x}={9}$$
Now taking,
$$\displaystyle{x}+{1}={0}$$
Add $$\displaystyle-{1}$$ on both sides,
$$\displaystyle{x}+{1}-{1}={0}-{1}$$
$$\displaystyle{x}=-{1}$$
Here, we neglect $$\displaystyle{x}=-{1}$$ because logarithm function cannot be negative.
Thus, the exact solution is $$\displaystyle{x}={9}.$$