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

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

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Answer 1 · 2016-12-29T12:15:18

$x = - 5$ $x=-5$

Explanation:

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

there are $2$ $2$ ways of solving this:

$1)$ $1)$
convert to logarithmic form:

$a^{m} = n \to {log}_{a} (n) = m$ $a^m=n -> log_a(n)=m$

$(\frac{1}{2}) x = 32 \to {log}_{\frac{1}{2}} (32) = x$ $(1/2)x=32 -> log_(1/2)(32)=x$

${log}_{\frac{1}{2}} (32) = {log}_{0.5} (32)$ $log_(1/2)(32)=log_0.5(32)$

enter into a calculator:

${log}_{0.5} (32) = - 5$ $log_0.5(32)= -5$

$x = - 5$ $x=-5$

$2)$ $2)$

laws of indices:

$a^{- m} = \frac{1}{a^{m}}$ $a^-m = 1/(a^m)$

${(a^{m})}^{n} = a^{m n}$ $(a^m)^n=a^(mn)$

convert $(\frac{1}{2})$ $(1/2)$ and $32$ $32$ to powers of $2$ $2$ :

$(\frac{1}{2}) = \frac{1}{2^{1}} = 2^{- 1}$ $(1/2) = 1/(2^1)=2^-1$

$32 = 2^{5}$ $32 = 2^5$

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

${(2^{- 1})}^{x} = 2^{5}$ $(2^-1)^x=2^5$

$2^{- 1 \cdot x} = 2^{5}$ $2^(-1 * x) = 2^5$

$- 1 \cdot x = 5$ $-1*x = 5$

$- x = 5$ $-x = 5$

divide by $- 1$ $-1$ :

$x = - 5$ $x = -5$

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1 Answer

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