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.
$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2} = x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^{x - 1}$
$(x^2 - 7x + 5)^{x^2-2x-15} = 1$
On
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
On
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 $2^{x-1}=2^{|x-3|+2}$
On
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 $$x^2-7x+5=1\\x^2-2x-15=0$$
On
For the first on, put like terms together.
$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2} = x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^{x - 1}$
$x^2 \cdot 2^{x + 1}-x^2 \cdot 2^{\lvert x - 3\rvert + 4} =2^{x - 1}-2 ^{\lvert x - 3\rvert + 2}$
$x^2(2^{x + 1}-2^{\lvert x - 3\rvert + 4})= 2^{x - 1}-2 ^{\lvert x - 3\rvert + 2}$
If $2^{x + 1}-2^{\lvert x - 3\rvert + 4}= 0$
Then $2^{x+1} = 2^{\lvert x - 3\rvert + 4}$
$\log_2 2^{x+1} = \log_2 2^{\lvert x - 3\rvert + 4}$
$x + 1 = |x -3| +4$
$x-3 = |x-3|$ which simply means $x \ge 3$.
But
$x^2(2^{x + 1}-2^{\lvert x - 3\rvert + 4})= 0 = 2^{x - 1}-2 ^{\lvert x - 3\rvert + 2}$
So $2^{x - 1}= 2 ^{\lvert x - 3\rvert + 2}$
$\log_2 2^{x - 1}= \log_2 2 ^{\lvert x - 3\rvert + 2}$
$x - 1 = |x - 3| + 2$
$x - 3 = |x-3|$ so ... again $x \ge 3$ will be a solution.
So thats one set of solutions $x \in [3,\infty)$.
If $2^{x + 1}-2^{\lvert x - 3\rvert + 4}\ne 0$ then $x < 3$ though then $|x - 3| = 3 -x$ and we have:
$x^2(2^{x + 1}-2^{\lvert x - 3\rvert + 4})= 2^{x - 1}-2 ^{\lvert x - 3\rvert + 2}$
$x^2(2^{x + 1}-2^{3-x + 4})= 2^{x - 1}-2 ^{3-x + 2}$
$x^2(2^{x+1} - 2^{7-x}) = 2^{x-1}-2^{5-x}$
$x^2 = \frac{2^{x-1}-2^{5-x}}{2^{x+1} - 2^{7-x}}$
$x^2 = \frac{2^{x-1}-2^{5-x}}{2^2(2^{x-1} - 2^{5-x})}$
$x^2 = \frac{1}{2^2}=1/4$
$x = \pm \frac 12$
So $x \in \{\pm 1/4\} \cup [3, \infty)$.
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Number 2 is .... a little clever.
If $b^c = 1$ then either
i) $c = 0$
ii) $b = 1$
iii) $b = -1$ and $c$ is an "even rational" (a rational number that when expressed as a ratio of two co-prime integers has a numerator divisible by two).
if i) $x^2 - 2x - 15 = 0$
$(x -5)(x+3) = 0$
so $x = 5$ or $x =- 3$
and we have $(x^2 - 7x + 5)^{x^2-2x-15} = (25 - 35 + 5)^{25-10 - 15} = (-5)^0 = 1$
or $(x^2 - 7x + 5)^{x^2-2x-15} = (9 + 21 + 5)^{9 + 6 - 15} = 35^0 = 1$
if ii)$x^2 - 7x + 5 = 1$
$x^2 - 7x + 4= 0$
$x = \frac{7\pm\sqrt{49 - 16}}{2}= \frac{7\pm\sqrt{33}}{2}$
if iii) $x^2 - 7x +5 = -1$
$x^2 -7x +6 = 0$
$(x - 1)(x - 6) = 0$
$x = 1, 6$
$1^2-2*1-15= -16$ is even and $6^2 - 2*6 - 15$ is odd.
So $(x^2 - 7x + 5)^{x^2-2x-15} = (1 - 7 + 5)^{1-2 - 15} = (-1)^{-16} = \frac 1{1^{16}} = 1$
[But $(x^2 - 7x + 5)^{x^2-2x-15} = (36 - 42 + 5)^{36 - 12 -15} = (-1)^{9} = -1 \ne 1$ ]
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All this is presuming we are only considering real numbers.
On
$$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2} = x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^{x - 1}$$
Let us distinguish two cases, $x\ge3$ and $x\le3$, to get rid of the absolute value.
$$x^2 \cdot 2^{x + 1} +2 ^{x-1} = x^2 \cdot 2^{x+1} + 2^{x - 1},$$ which is an identity !
$$x^2 \cdot 2^{x + 1} +2 ^{5-x} = x^2 \cdot 2^{7-x} + 2^{x - 1}$$ 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 $$b=0\to x=-3,5.$$
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.