Powerful Exponents

Algebra Level 3

Let a a and b b be real numbers such that 6 0 a = 3 60^a = 3 and 6 0 b = 5 60^b= 5 . Find

1 2 1 a b 2 ( 1 b ) \Large 12^{\frac{1-a-b}{2(1-b)}}


The answer is 2.

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1 solution

Chew-Seong Cheong
Dec 31, 2016

{ 6 0 a = 3 a log 60 = log 3 a = log 3 log 60 6 0 b = 5 b log 60 = log 5 b = log 5 log 60 \begin{cases} 60^a = 3 & \implies a \log 60 = \log 3 & \implies a = \dfrac {\log 3}{\log 60} \\ 60^b = 5 & \implies b \log 60 = \log 5 & \implies b = \dfrac {\log 5}{\log 60} \end{cases}

x = 1 2 1 a b 2 ( 1 b ) log x = 1 a b 2 ( 1 b ) log 12 log x log 12 = 1 log 3 log 60 log 5 log 60 2 ( 1 log 5 log 60 ) = log 60 log 3 log 5 2 ( log 60 log 5 ) = log 60 3 × 5 2 × log 60 5 = log 4 2 × log 12 log x = log 4 × log 12 2 × log 12 = 2 log 2 2 = log 2 x = 2 \begin{aligned} x & = 12^{\frac {1-a-b}{2(1-b)}} \\ \log x & = \frac {1-a-b}{2(1-b)} \log 12 \\ \frac {\log x}{\log 12} & = \frac {1-\frac {\log 3}{\log 60} - \frac {\log 5}{\log 60}}{2\left(1-\frac {\log 5}{\log 60}\right)} \\ & = \frac {\log 60 - \log 3 - \log 5}{2(\log 60 - \log 5)} \\ & = \frac {\log \frac {60}{3 \times 5}}{2 \times \log \frac {60}5} \\ & = \frac {\log 4}{2 \times \log 12} \\ \implies \log x & = \frac {\log 4 \times \log 12}{2 \times \log 12} = \frac {2 \log 2}2 = \log 2 \\ \implies x & = \boxed{2} \end{aligned}

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