I'd appreicate any assistance in answering this questionHeadbang)
3^x+1=6^x-1. The answer needs to be rounded to three decimal places. Thanks in advance.
I assume the equation is $\displaystyle 3^{x + 1} = 6^{x - 1}
$
Take logs:
$\displaystyle (x + 1) \log 3 = (x - 1) \log 6$
Rearrange:
$\displaystyle \frac{x + 1}{x - 1} = \frac{\log 6}{\log 3}$
$\displaystyle \frac{x - 1}{x - 1} + \frac{2}{x - 1} = \frac{\log 6}{\log 3}$
$\displaystyle 1 + \frac{2}{x - 1} = \frac{\log 6}{\log 3}$
$\displaystyle \frac{2}{x - 1} = \frac{\log 6}{\log 3} - 1$
Can you solve it now?