Solve $6^{x+8} = 4^{x-1}$ I tried doing $log_6\left(6^{x+8}\right) = log_6{4^{x-1}}$
I got stuck, and I don't think that was the right route.
 A: The OP's start was fine.  We rewrite the left side as $x+8$, and pull the exponent out of the right side, to get $$x+8=(x-1)\log_64$$
We expand the right side to get $$x+8=x(\log_64)-\log_64$$
then subtract $8+x\log_64$ from both sides, to get
$$x-x\log_64=-8-\log_64$$
We now factor out $x$ to get $$x(1-\log_64)=-8-\log_64$$
Lastly, we divide by $1-\log_64$ to get $$x=\frac{-8-\log_64}{1-\log_64}$$
A: It doesn't necessarily matter which base you are in, you can solve this as follows $$6^{x+8}=4^{x-1}\iff \log(6^{x+8})=\log(4^{x-1})$$ then, by rules of the logarithm, $$(x+8)\log 6=(x-1)\log 4$$
now, by the distributive property for real numbers,
$$(x+8)\log 6=(x-1)\log 4 \iff x\log 6 + 8\log 6 = x\log 4 -\log4$$
and so
$$x\log6-x\log 4=-8\log 6 - \log4$$
again by the distributive property
$$x(\log6-\log4)=-8\log 6 - \log4$$
and since $(\log 6 - \log 4)\neq 0$, we can divide both sides by it to achieve:
$$x=\frac{-8\log 6 - \log 4}{\log6-\log4}$$
A: *

*Remove the logs of exponents


$$(6x+8)*\log_6(6)=(x-1)*\log_6(4)$$


*Rewrite bases


$$(6x+8)*\frac{\log(6)}{\log(6)}=(x-1)*\frac{\log(4)}{\log(6)}$$


*Since $\frac{\log(6)}{\log(6)} = 1$,


$$(6x+8)=(x-1)*\frac{\log(4)}{\log(6)}$$


*Thus, dividing $\frac{(6x+8)}{(x-1)}$:


$$\frac{(6x+8)}{(x-1)}=\frac{\log(4)}{\log(6)}$$


*Then, after further solving the equation, we get the value of $x$:


$$x=-\frac{\log(2592)}{\log(108)}$$
