Homework help: Exponential and Logarithmic Equations I am kinda stuck with the last two equations. Just need a kick/hint in the right direction to get it done.
My question is: How to simplify them to some base form from which I can get the x value. I want to know the method.
$8^{x+3}=6\cdot 2^x+4\cdot 4^{x+2}$
$2\cdot\ln\left(x\right)-\ln\left(x-5\right)=\ln\left(x+1\right)$
 A: For the equation
$$8^{x + 3} = 6 \cdot 2^x + 4 \cdot 4^{x + 2}$$
use $2$ as the common base.
\begin{align*}
8^{x + 3} & = 6 \cdot 2^x + 4 \cdot 4^{x + 2}\\
(2^3)^{x + 3} & = 6 \cdot 2^x + 4 \cdot (2^2)^{x + 2}\\
2^{3x + 9} & = 6 \cdot 2^x + 4 \cdot 2^{2x + 4}\\
2^9 \cdot 2^{3x} & = 6 \cdot 2^x + 4 \cdot 2^4 \cdot 2^{2x}\\
512 \cdot 2^{3x} & = 6 \cdot 2^x + 64 \cdot 2^{2x}\\
256 \cdot 2^{3x} & = 3 \cdot 2^x + 32 \cdot 2^{2x}
\end{align*} 
Let $u = 2^x$ to obtain
$$256u^3 = 3u + 32u^2$$
and note that $u = 2^x > 0$ for every real number $x$, so $u = 0$ is not a solution of the cubic.
For the equation 
$$2\ln x - \ln (x - 5) = \ln (x + 1)$$
we obtain
\begin{align*}
\ln x^2 - \ln (x - 5) & = \ln (x + 1)\\
\ln\left(\frac{x^2}{x - 5}\right) & = \ln (x + 1)\\
e^{\ln\left(\frac{x^2}{x - 5}\right)} & = e^{\ln (x + 1)}\\
\frac{x^2}{x - 5} & = x + 1
\end{align*}
Keep in mind that $\ln x$ is only defined when $x > 0$, $\ln (x - 5)$ is only defined when $x - 5 > 0$, and $\ln (x + 1)$ is only defined when $x + 1 > 0$. Your final answer must satisfy all three of those restrictions. 
