Equation which has to be solved with logarithms I need some help how to solve these equations for $x$. I think I have to use logarithms but still not sure how to do it and would be really grateful if someone could explain me.
$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2}
 = x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^{x - 1}$
$(x^2 - 7x + 5)^{x^2-2x-15} = 1$
 A: Hint: if
\begin{equation*}
  (x^2 - 7x + 5)^{\color{blue}{x^2 - 2x - 15}} = 1,
\end{equation*}
what do you think $\color{blue}{x^2 - 2x - 15}$ is equal to?
This is merely one of the three possibilities. For all of them, please check out fleablood's answer.
A: 1) Making $2^{x-1}=a$ and $2^{|x-3|+2}=b$ you have $$4ax^2+b=4bx^2+a\iff(4x^2-1)(a-b)=0$$ This gives $x=\frac 12$ and $x\ge3$
2) You have two independent possibilities $$x^2-7x+5=1\\x^2-2x-15=0$$
A: $$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2}
 = x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^{x - 1}$$
Let us distinguish two cases, $x\ge3$ and $x\le3$, to get rid of the absolute value.


*

*$x\ge3$:


$$x^2 \cdot 2^{x + 1} +2 ^{x-1}
 = x^2 \cdot 2^{x+1} + 2^{x - 1},$$ which is an identity !


*$x\le3$:


$$x^2 \cdot 2^{x + 1} +2 ^{5-x}
 = x^2 \cdot 2^{7-x} + 2^{x - 1}$$
which we rewrite
$$\left(2x^2-\frac12\right)2^x=2^6\left(2x^2-\frac12\right)2^{-x},$$
so that $$x=\pm\frac12\text{ or }x=3.$$

$$(x^2 - 7x + 5)^{x^2-2x-15} = 1$$
$a^b=1$ when $a=1$ or $a=-1\land\text{even}(b)$ or $b=0$, so
$$a=0\to x=\frac{7\pm\sqrt33}2,$$ 
$$a=-1\to x=1\text{ or 6},$$ where $6$ must be rejected as it yields an odd exponent, and
$$b=0\to x=-3,5.$$
A: For the second one you should check when 
$$x^2-2x-15=0$$ 
or $$x^2-7x+5=\pm 1$$
Also for $x^2-7x+5=- 1$ you have to chekc whether $x^2-2x-15$ is even
A: For the first one write $a=x^2,b=2^{x-1},c={|x-3|+2}$ then you get 
$$4ab+c=4ac+b$$ or
$$b(4a-1)=c(4a-1)$$
Which means $4x^2=1$ or $2^{x-1}=2^{|x-3|+2}$
