Question

solve for x.
\\(3^{-2x} = 17^{x - 5}\\)
write the exact answer using either base-10 or base-\\(e\\) logarithms.
\\(x = \square\\)

Explanation:

Step1: Take log on both sides

Take the natural logarithm (ln) of both sides of the equation \(3^{-2x} = 17^{x - 5}\). Using the logarithm power rule \(\ln(a^b)=b\ln(a)\), we get:
\(-2x\ln(3)=(x - 5)\ln(17)\)

Step2: Expand the right - hand side

Expand the right - hand side of the equation: \(-2x\ln(3)=x\ln(17)-5\ln(17)\)

Step3: Move all x - terms to one side

Move the \(x\ln(17)\) term to the left - hand side: \(-2x\ln(3)-x\ln(17)=- 5\ln(17)\)
Factor out \(x\) from the left - hand side: \(x(-2\ln(3)-\ln(17))=-5\ln(17)\)

Step4: Solve for x

Divide both sides by \(-2\ln(3)-\ln(17)\) to solve for \(x\). We can also factor out a negative sign from the denominator:
\(x=\frac{-5\ln(17)}{-2\ln(3)-\ln(17)}=\frac{5\ln(17)}{2\ln(3)+\ln(17)}\)
(We could also use common logarithms (log base 10) instead of natural logarithms. The process would be similar: take \(\log\) of both sides, apply the power rule, and then solve for \(x\). For example, starting with \(\log(3^{-2x})=\log(17^{x - 5})\), we get \(-2x\log(3)=(x - 5)\log(17)\), and then follow the same steps of expanding, moving terms, and factoring to solve for \(x\).)

Answer:

\(x = \frac{5\ln(17)}{2\ln(3)+\ln(17)}\) (or an equivalent expression using base - 10 logarithms, e.g., \(x=\frac{5\log(17)}{2\log(3)+\log(17)}\))