The great log

Algebra Level 2

( 8 1 1 log ⁡ 5 9 + 3 3 log ⁡ 6 3 409 ) ( ( 7 ) 2 log ⁡ 25 7 − ( 125 ) log ⁡ 25 6 ) = ? \Large \left( \frac{81^{\frac{1}{\log_{5}9}}+3^{\frac{3}{\log_{\sqrt{6}}3}}}{409}\right) \left((\sqrt{7})^{\frac{2}{\log_{25}7}}-(125)^{\log_{25}6}\right)= \, ?


The answer is 1.

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Ayushman Chahar by Ayushman Chahar , 21, India

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2 solutions

Akshat Sharda
Jan 2, 2016

= ( 8 1 1 log ⁡ 5 9 + 3 3 log ⁡ 6 3 409 ) ( ( 7 ) 2 log ⁡ 25 7 − ( 125 ) log ⁡ 25 6 ) = ( 9 2 log ⁡ 9 5 + 3 3 log ⁡ 3 6 409 ) ( 7 log ⁡ 7 25 − 5 3 2 log ⁡ 5 6 ) = ( 25 + 6 6 409 ) ( 25 − 6 6 ) = ( 25 ) 2 − ( 6 6 ) 2 409 = 409 409 = 1 =\left( \frac{81^{\frac{1}{\log_{5}9}}+3^{\frac{3}{\log_{\sqrt{6}}3}}}{409}\right) \left((\sqrt{7})^{\frac{2}{\log_{25}7}}-(125)^{\log_{25}6}\right) \\ =\left(\frac{9^{2\log_{9}{5}}+3^{3\log_{3}\sqrt{6}}}{409}\right) \left(7^{\log_{7}{25}}-5^{\frac{3}{2}\log_{5}6}\right) \\ = \left( \frac{25+6\sqrt{6}}{409}\right) \left(25-6\sqrt{6}\right) \\ = \frac{(25)^2-(6\sqrt{6})^2}{409}=\frac{409}{409}=\boxed{1}

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Mohit Bhavsar
Jan 3, 2016

If you want to save your time, you should know that if there are such big questions, then, the answer would always be 0,1,2 or 3. So simple I didn't even calculate anything and i just tried typing '1' and see, it went correct.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

But its not the main way to solve a particular question. The main thing is that you must know the solution to solve the problem. dont use hit and trial.

Ayushman Chahar - 5 years, 5 months ago

I did the same :')

Sagar Varsani - 5 years, 5 months ago

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