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.
$8^{x+3}=6\cdot 2^x+4\cdot 4^{x+2}$
$2\cdot\ln\left(x\right)-\ln\left(x-5\right)=\ln\left(x+1\right)$
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.