Solve the equation: $2^{2x+1}=\left(\frac{1}{32}\right)^x$ Having trouble with this problem:
$$2^{2x+1}=\frac{1}{32^x}$$
Do I need to set the exponents equal to each other?
 A: You you can equate exponents only if you have the same base.
$$2^{2x+1} = 2^{x\log_2{\frac{1}{32}}}$$
so then we have
$$2x+1 = x\log_2{\frac{1}{32}}=x\log_2{2^{-5}}=-5x$$
and
$$2x + 1 = -5x$$
so
$$x = \frac{-1}{7}$$
You could have also taken the $\log_2$ of both sides and it would have given you the same answer.
A: Notice, here is easier method without using logarithms $$2^{2x+1}=\left(\frac{1}{32}\right)^{x}$$
$$ 2^{2x+1}=\left(\frac{1}{2^5}\right)^{x}=\frac{1}{2^{5x}}$$
$$2^{2x+1}\cdot 2^{5x}=1$$
$$2^{7x+1}=2^0$$
comparing the powers of base $2$ on both the sides, one should get
$$7x+1=0$$
$$\color{red}{x=-\frac{1}{7}}$$
A: $2^{2x+1}=\dfrac{1}{32^x}\iff$
$32^x\cdot2^{2x+1}=1\iff$
$(2^5)^x\cdot2^{2x+1}=1\iff$
$2^{5x}\cdot2^{2x+1}=1\iff$
$2^{5x+2x+1}=1\iff$
$2^{7x+1}=1\iff$
$7x+1=\log_21\iff$
$7x+1=0\iff$
$7x=-1\iff$
$x=-\dfrac17$
A: Just logarithm both sides to get
$$
(2x+1) \log(2) = x \log(1/32) \Rightarrow \\
(2 \log(2) - \log(1/32)) x + \log(2)  = 0 \Rightarrow \\
x = -\log(2) / \log(128)
$$
where $\log(x^y) = y \log(x)$ and $\log(x) + \log(y) = \log(x\,y)$ was used. At this point it is clear that using base $2$ for the logarithm simplifies the calculation:
$$
x = -1/7
$$
