Solve the equation: $2^{2x+1}=\left(\frac{1}{32}\right)^x$

Question

Having trouble with this problem:

$$2^{2x+1}=\frac{1}{32^x}$$

Do I need to set the exponents equal to each other?

Answer 1

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.

Answer 2

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 $$7x+1=0$$ $$\color{red}{x=-\frac{1}{7}}$$

Answer 3

$2^{2x+1}=\dfrac{1}{32^x}\iff$

$32^x\cdot2^{2x+1}=1\iff$

$(2^5)^x\cdot2^{2x+1}=1\iff$

$2^{5x}\cdot2^{2x+1}=1\iff$

$2^{5x+2x+1}=1\iff$

$2^{7x+1}=1\iff$

$7x+1=\log_21\iff$

$7x+1=0\iff$

$7x=-1\iff$

$x=-\dfrac17$

Answer 4

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: $$ x = -1/7 $$