Solve for x.
pi^7x-1 = e^8x
I think I need to take the natural log of both sides. but I don't know what to do after this step-
The answer should be exact, which means it should be written with ln and pi and all that
next time use parentheses around the entire exponent to make your post clear.
$\displaystyle \pi^{7x-1} = e^{8x}$
$\displaystyle \ln(\pi^{7x-1}) = \ln(e^{8x})
$
$\displaystyle (7x-1)\ln{\pi} = 8x$
$\displaystyle 7x \cdot \ln{\pi} - \ln{\pi} = 8x$
$\displaystyle 7x \cdot \ln{\pi} - 8x = \ln{\pi}$
$\displaystyle x(7\ln{\pi} - 8) = \ln{\pi}$
$\displaystyle x = \frac{\ln{\pi}}{7\ln{\pi} - 8}$