2010 AMC 12B Problem 12

Attempt Problem 12 of the 2010 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12B solutions, or check the answer key.

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12.

For what value of xx does log⁡2x+log⁡2x+log⁡4(x2)+log⁡8(x3)+log⁡16(x4)=40? \begin{aligned} &\log_{\sqrt2}\sqrt{x}+\log_2 x \\ &\quad {}+\log_4\left(x^2\right)+\log_8\left(x^3\right) \\ &\quad {}+\log_{16}\left(x^4\right)=40? \end{aligned}
88
1616
3232
256256
10241024
Answer: D
Concepts:logarithm
Small hint:
Convert every term to base 2;2; let L=log⁡2xL=\log_2 x
Big hint:
Each of the five terms simplifies to LL
Solution:
Let L=log⁡2x.L=\log_2 x. Converting each term to base 2:2:
log⁡2x=L212=L,\log_{\sqrt2}\sqrt{x}=\dfrac{\frac{L}{2}}{\frac{1}{2}}=L, log⁡2x=L,\log_2 x=L, log⁡4x2=2L2=L,\log_4 x^2=\dfrac{2L}{2}=L, log⁡8x3=3L3=L,\log_8 x^3=\dfrac{3L}{3}=L, and log⁡16x4=4L4=L.\log_{16}x^4=\dfrac{4L}{4}=L.
The equation becomes 5L=40,5L=40, so L=8L=8 and x=28=256.x=2^8=256.
Thus, the correct answer is D.
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