Break down the problem into these 2 equations.
\[{5}^{x-2}=1\]
\[{5}^{x-2}=FINDth()\]
Solve the 1st equation: \({5}^{x-2}=1\).
Use Definition of Common Logarithm: \({b}^{a}=x\) if and only if \(log_b(x)=a\).
\[x-2=\log_{5}{1}\]
Use Change of Base Rule: \(\log_{b}{x}=\frac{\log_{a}{x}}{\log_{a}{b}}\).
\[x-2=\frac{\log{1}}{\log{5}}\]
Use Rule of One: \(\log{1}=0\).
\[x-2=\frac{0}{\log{5}}\]
Simplify \(\frac{0}{\log{5}}\) to \(0\).
\[x-2=0\]
Add \(2\) to both sides.
\[x=2\]
\[x=2\]
Solve the 2nd equation: \({5}^{x-2}=FINDth()\).
Use Definition of Common Logarithm: \({b}^{a}=x\) if and only if \(log_b(x)=a\).
\[x-2=\log_{5}{(FINDth)}\]
Add \(2\) to both sides.
\[x=\log_{5}{(FINDth)}+2\]
\[x=\log_{5}{(FINDth)}+2\]
Collect all solutions.
\[x=2,\log_{5}{(FINDth)}+2\]
x=2,log(5,FINDt*h*)+2
