How do you solve $\frac{1}{32^{2 x}} = 64$ $1/32^(2x)=64$ ?

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Answer 1 · 2016-04-25T15:52:46

$x = - \frac{3}{5}$ $x=-3/5$

Using properties of exponents we are going to rewrite everything in terms of base $2$ $2$ .

$\frac{1}{{(32^{2})}^{x}} = 64$ $1/(32^2)^x=64$

$\frac{1}{{({(2^{5})}^{2})}^{x}} = 64$ $1/((2^5)^2)^x=64$

$\frac{1}{2^{10 x}} = 2^{6}$ $1/2^(10x)=2^6$

$2^{- 10 x} = 2^{6}$ $2^(-10x)=2^6$

Since we are now in the same base we can do the following

$- 10 x = 6$ $-10x=6$

$x = \frac{6}{- 10} = - \frac{3}{5}$ $x=6/-10=-3/5$

How do you solve 1322x=641/32^(2x)=64?