How do you solve ${0.5}^{x} = 16^{2}$ $0.5^x=16^2$ ?

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How do you solve ${0.5}^{x} = 16^{2}$ $0.5^x=16^2$ ?

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Answer 1 · 2016-09-08T16:02:27

$x = - 8$ $x = -8$

Explanation:

When you are working with equations with indices, try to

either $\to make the bases the same$ $rarr " make the bases the same"$

or $\to make the indices the same$ $rarr " make the indices the same"$

$0.5 = \frac{1}{2}$ $0.5 = 1/2$ This is a better form to use, because you should recognise that 16 is one of powers of 2.......

$\to 2^{4} = 16 and 2^{8} = 16^{2}$ $rarr2^4 = 16 and 2^8 = 16^2$

${(\frac{1}{2})}^{x} = 16^{2}$ $(1/2)^x = 16^2$

${(\frac{1}{2})}^{x} = 2^{8}$ $(1/2)^x = 2^8$

$2^{- x} = 2^{8} \to one of the index laws: {(\frac{a}{b})}^{m} = {(\frac{b}{a})}^{- m}$ $2^-x = 2^8 " " rarr "one of the index laws: "(a/b)^m = (b/a)^-m$

Now the bases are the same so we have:

$- x = 8 \Leftrightarrow x = - 8$ $-x = 8 hArr x = -8$

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How do you solve ${0.5}^{x} = 16^{2}$ $0.5^x=16^2$ ?

1 Answer

Explanation:

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