Question

solve for x.
$4^{9x} = 17^{-x + 9}$
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 \(4^{9x} = 17^{-x + 9}\). Using the logarithm power rule \(\ln(a^b)=b\ln(a)\), we get:
\(9x\ln(4)=(-x + 9)\ln(17)\)

Step2: Expand the right - hand side

Expand the right - hand side of the equation:
\(9x\ln(4)=-x\ln(17)+9\ln(17)\)

Step3: Move terms with x to the left

Add \(x\ln(17)\) to both sides of the equation to get all terms with \(x\) on the left - hand side:
\(9x\ln(4)+x\ln(17)=9\ln(17)\)

Step4: Factor out x

Factor out \(x\) from the left - hand side of the equation:
\(x(9\ln(4)+\ln(17)) = 9\ln(17)\)

Step5: Solve for x

Divide both sides of the equation by \((9\ln(4)+\ln(17))\) to solve for \(x\):
\(x=\frac{9\ln(17)}{9\ln(4)+\ln(17)}\) (We could also use base - 10 logarithms. If we use \(\log\) (base - 10), the steps are similar: take \(\log\) of both sides, apply the power rule, and then solve for \(x\), and we would get \(x = \frac{9\log(17)}{9\log(4)+\log(17)}\))

Answer:

\(x=\frac{9\ln(17)}{9\ln(4)+\ln(17)}\) (or \(x=\frac{9\log(17)}{9\log(4)+\log(17)}\))