How do you solve ${1.8}^{x} = {(\frac{1}{32})}^{x - 2}$ $1.8^x=(1/32)^(x-2)$ ?

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How do you solve ${1.8}^{x} = {(\frac{1}{32})}^{x - 2}$ $1.8^x=(1/32)^(x-2)$ ?

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Answer 1 · 2016-03-06T16:07:20

1.71, nearly.

Explanation:

log ( $a^{m}$ $a^m$ ) = m log a
Equate logarithms (of any base ) of the expressions, on both the sides.
x log 1.8 = - ( x $-$ $-$ 2) log 32
x = $2 \frac{log 32}{log 1.8 + log 32}$ $2log32/(log1.8+log32)$

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How do you solve ${1.8}^{x} = {(\frac{1}{32})}^{x - 2}$ $1.8^x=(1/32)^(x-2)$ ?

1 Answer

Explanation:

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