Exponentiation Problem

Can anyone please help me in solving this problem?

$60^{a} = 3$

$60^{b} = 5$

$12^{\frac{1 - a - b}{2(1 - b)}} = ?$

I am just unable to solve it. So, any help would be highly appreciated.

Thanks.

#Algebra

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

$a=\dfrac{\log3}{\log{60}}$ . $b=\dfrac{\log5}{\log{60}}$ .

$12^{\frac{1-(a+b)}{2(1-b)}}$ $=12^{\frac{1-\log15/\log{60}}{2(1-\log5/{60})}}=$ $12^{\frac{\log4/\log{60}}{2(\log{12}/\log{60})}}=$ $12^{\log2/\log{12}}=12^{\log_{12}{2}}=2$ .

Note:

$\log a+\log b=\log{ab}$

$\log a-\log b=\log{a/b}$

$\log a^2=2 \log a$

$a^{\log_a b}=b$ .

Here, we take $1=\log{60}/\log{60}$ .

A Former Brilliant Member - 5 years, 2 months ago

Thank you very much.☺ Helped me understand a lot.😃😃😃

Nilabha Saha - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$