Logging x x

Algebra Level 2

Solve for x x :

log 2 x 216 = x . \large\log_{2x}{216}=x.


Source: AHSME 1960.


The answer is 3.

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2 solutions

Hobart Pao
Aug 28, 2016

( 2 x ) x = 216 (2x)^x = 216 2 x x x = 2 3 3 3 2^x x^x = 2^3 3^3 x = 3 x = \boxed{3}

Chew-Seong Cheong
Aug 28, 2016

log 2 x 216 = x log 2 ( 2 3 3 3 ) log 2 ( 2 x ) = x 3 + 3 log 2 3 1 + log 2 x = x Multiplying both sides by 1 + log 2 x 3 + 3 log 2 3 = x + x log 2 x Noting the positions of x and 3 on both sides x = 3 \begin{aligned} \log_{2x} 216 & = x \\ \frac {\log_2 (2^33^3)}{\log_2 (2x)} & = x \\ \frac {3+3\log_2 3}{1+\log_2 x} & = x & \small \color{#3D99F6} \text{Multiplying both sides by }1+\log_2 x \\ {\color{#3D99F6}3} + {\color{#3D99F6}3}\log_2 {\color{#3D99F6}3} & = {\color{#3D99F6}x} + {\color{#3D99F6}x}\log_2 \color{#3D99F6}x & \small \color{#3D99F6} \text{Noting the positions of } x \text{ and 3 on both sides} \\ \implies \color{#3D99F6}{x} & = \boxed{\color{#3D99F6}{3}} \end{aligned}

Sir, explain the 4th statement.

A Former Brilliant Member - 3 years, 3 months ago

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I have added notes to explain. Hope they help.

Chew-Seong Cheong - 3 years, 3 months ago

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Can't we do it without judging the positions? rigorously?

A Former Brilliant Member - 3 years, 3 months ago

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@A Former Brilliant Member I don't know, but this is rigorous enough for me.

Chew-Seong Cheong - 3 years, 3 months ago

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@Chew-Seong Cheong I also got the logic. What is true for 3 + log 2 3 3 + \log_{2} 3 is also true for x + x log 2 x x + x \log_{2} x . But can't we add up few more lines to mathematically show that x = 3 x = 3 . That's what I wanted to know. Thank you.

A Former Brilliant Member - 3 years, 3 months ago

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@A Former Brilliant Member Just look at it as mathematical induction. We can say that If f ( x ) = x + x log 2 x f(x) = x + x \log_2 x , then f ( 3 ) = 3 + 3 log 2 3 f(3) = 3+3\log_2 3 . Therefore, x = 3 x=3 . But I don't think it is all necessary.

Chew-Seong Cheong - 3 years, 3 months ago

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@Chew-Seong Cheong Thank you for your time and attention. I hope you will be successful in sharing your indulgence.

A Former Brilliant Member - 3 years, 3 months ago

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@A Former Brilliant Member Thanks. Hope you having much fun learning.

Chew-Seong Cheong - 3 years, 3 months ago

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