How do you solve 3*18^(n+8)=1?

Aug 24, 2016

$n = - \left(8 + \frac{1}{{\log}_{3} 2 + 2}\right)$

Explanation:

$3 \cdot {18}^{n + 8} = 1$. Applying the $\log$ transformation to both sides

${\log}_{e} 3 + \left(n + 8\right) {\log}_{e} 18 = 0$ or
$n = - \frac{8 {\log}_{e} 18 + {\log}_{e} 3}{{\log}_{e} 18} = - 8 - {\log}_{e} \frac{3}{{\log}_{e} 18}$ then

n = -8-log_e3/(log_e2+2log_e 3) = -8-1/(log_e2/log_e3+2

Finally

$n = - \left(8 + \frac{1}{{\log}_{3} 2 + 2}\right)$