At this point we need to take the logarithm of both sides so we can use logarithm properties to get the \(t\) out of the exponent. It doesn’t matter which logarithm we use, but if we want a decimal value for the answer it will need to be one that we can work with. For this solution we’ll use the natural logarithm.

Upon taking the logarithm we then need to use logarithm properties to get the \(t\)’s out of the exponent at which point we can solve for \(t\). Here is the rest of the work for this problem,

\[\begin{align*}\ln \left( {{{17}^{t - 2}}} \right) & = \ln \left( 5 \right)\\ \left( {t - 2} \right)\ln \left( {17} \right) & = \ln \left( 5 \right)\\ t - 2 & = \frac{{\ln \left( 5 \right)}}{{\ln \left( {17} \right)}}\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{2 + \frac{{\ln \left( 5 \right)}}{{\ln \left( {17} \right)}} = 2.5681}}\end{align*}\]

Depending upon your preferences either the exact or decimal solution can be used. Also note that if you had used, say the common logarithm, you would get exactly the same answer.