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Algebra Level 3

( ( log ⁡ 2 9 ) 2 ) 1 log ⁡ 2 ( log ⁡ 2 9 ) × ( 7 ) 1 log ⁡ 4 7 = ? \Large \left((\log_{2}9)^{2}\right)^{\frac {1}{\log_{2}(\log_{2}9)}} \times \left(\sqrt{7}\right)^{\frac {1}{\log_{4}7}} =\ ?


The answer is 8.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

( ( log ⁡ 2 9 ) ) 2 log ⁡ log ⁡ 2 9 2 × ( 7 ) 1 2 × log ⁡ 7 4 ((\log_{2}9))^{2\log_{\log_29}2} \times (7)^{\frac {1}{2}\times \log_{7}4}

.(Since , log ⁡ a b = 1 log ⁡ b a \log_{a}b = \dfrac {1}{\log_{b}a} )

= ( ( log ⁡ 2 9 ) ) log ⁡ log ⁡ 2 9 2 2 × ( 7 ) log ⁡ 7 4 1 2 = ((\log_{2}9))^{\log_{\log_29}2^{2}} \times (7)^{\log_{7}4^{\frac {1}{2}}}

= 2 2 × 4 1 2 = 4 × 2 = 2^{2}\times 4^{\frac{1}{2}} = 4\times 2

(Since, a l o g a b = b a^{log_{a}b} = b )

A N S W E R : 8 ANSWER:\boxed{8}

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Chew-Seong Cheong
Nov 14, 2018

Similar solution with @Niraj Sawant 's

X = ( ( log ⁡ 2 9 ) 2 ) 1 log ⁡ 2 ( log ⁡ 2 9 ) ( 7 ) 1 log ⁡ 4 7 = ( ( log ⁡ 2 9 ) 1 log ⁡ 2 ( log ⁡ 2 9 ) ) 2 ( 7 1 2 ) 1 log ⁡ 4 7 Note that log ⁡ log ⁡ 2 9 2 = log ⁡ 2 2 log ⁡ 2 ( log ⁡ 2 9 ) = 1 log ⁡ 2 ( log ⁡ 2 9 ) = ( ( log ⁡ 2 9 ) log ⁡ log ⁡ 2 9 2 ) 2 ( 7 log ⁡ 7 4 ) 1 2 Similarly 1 log ⁡ 4 7 = log ⁡ 7 4 = 2 2 × 4 1 2 Also that a log ⁡ a b = b = 8 \large {\begin{aligned} X & = \left((\log_{2}9)^{2}\right)^{\frac {1}{\log_{2}(\log_{2}9)}} \left(\sqrt{7}\right)^{\frac {1}{\log_{4}7}} \\ & = \left((\log_{2}9)^{\color{#3D99F6}\frac {1}{\log_{2}(\log_{2}9)}}\right)^2 \left(7^\frac 12 \right)^{\color{#3D99F6}\frac {1}{\log_{4}7}} & \small \color{#3D99F6} \text{Note that }\log_{\log_2 9} 2 = \frac {\log_2 2}{\log_2 (\log_2 9)} = \frac 1{\log_2 (\log_2 9)} \\ & = \left((\log_{2}9)^{\color{#3D99F6}\log_{\log_2 9} 2}\right)^2 \left(7^{\color{#3D99F6}\log_7 4}\right)^\frac 12 & \small \color{#3D99F6} \text{Similarly } \frac 1{\log_4 7} = \log_7 4 \\ & = 2^2 \times 4^\frac 12 & \small \color{#3D99F6} \text{Also that }a^{\log_a b} = b \\ & = \boxed 8 \end{aligned}}

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Thank you, your solution is more elaborate. Can you tell me how to get the size of parenthesis bigger in Latex? I would also like to know how to use integral notation in Latex

A Former Brilliant Member - 2 years, 6 months ago

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I actually edited your problem here. Use \ [ \ ] instead of \ ( \ ) it will solve a lot of formatting issues. Use \left( \right), it will automatically adjust to the size needed. But don't use it for one layer because it will make the bracket slightly bigger. Use backslash for most functions for example, integration \int, summation \sum ∑ \sum , \tan \sin \cos \csc \sec \cot product \prod ∏ \prod , \infty ∞ \infty , \alpha α \alpha , \beta β \beta , \gamma γ \gamma , \Gamma Γ \Gamma . If you use \ [ \ ] just use \int, it will be centralized as below. ∫ 0 ∞ f ( x ) d x = ∫ 0 ∞ f ( 1 x ) x 2 d x \int_0^\infty f(x) \ dx = \int_0^\infty \frac {f\left(\frac 1x\right)}{x^2} dx if you use \ ( \ ) \int_0^\frac \pi 2 is as ∫ 0 π 2 \int_0^\frac \pi 2 . If you want proper size put \displaystyle in front of \int as ∫ 0 π 2 \displaystyle \int_0^\frac \pi 2 . No need to use braces { } if only one character, for example \frac \pi 2 π 2 \frac \pi 2 \binom nk ( n k ) \binom nk put a d in front for display mode \dfrac 34 3 4 \dfrac 34 \dbinom nk ( n k ) \dbinom nk .

You can actually see the LaTex code in this website by placing mouse cursor on the formulas. Try it.

Chew-Seong Cheong - 2 years, 6 months ago

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Thanks a lot:)

A Former Brilliant Member - 2 years, 6 months ago

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