Difference between revisions of "2024 AMC 12B Problems/Problem 8"
Kafuu chino (talk | contribs) (→Solution 1) |
Sourodeepdeb (talk | contribs) |
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\end{align*} | \end{align*} | ||
so <math>\boxed{\textbf{(C) }36}</math> | so <math>\boxed{\textbf{(C) }36}</math> | ||
| + | |||
| + | ==Solution 2 (Change of Base)== | ||
| + | \begin{align*} | ||
| + | \frac{\log_2x \cdot \log_3x}{\log_2x+\log_3x} &= 2 \\[6pt] | ||
| + | \log_2x \cdot \log_3x &= 2(\log_2x+\log_3x) \\[6pt] | ||
| + | \log_2x \cdot \log_3x &= 2\log_2x + 2\log_3x \\[6pt] | ||
| + | \frac{\log x}{\log 2} \cdot \frac{\log x}{\log 3} &= 2\frac{\log x}{\log 2} + 2\frac{\log x}{\log 3} \\[6pt] | ||
| + | \frac{(\log x)^2}{\log 2 \cdot \log 3} &= \frac{2\log x \cdot \log 3 + 2\log x \cdot \log 2}{\log 2 \cdot \log 3} \\[6pt] | ||
| + | (\log x)^2 &= 2\log x \cdot \log 3 + 2\log x \cdot \log 2 \\[6pt] | ||
| + | (\log x)^2 &= 2\log x(\log 2 + \log 3) \\[6pt] | ||
| + | \log x &= 2(\log 2 + \log 3) \\[6pt] | ||
| + | x &= 10^{2(\log 2 + \log 3)} \\[6pt] | ||
| + | x &= (10^{\log 2} \cdot 10^{\log 3})^2 \\[6pt] | ||
| + | x &= (2 \cdot 3)^2 = 6^2 = \boxed{\textbf{(C) }36} | ||
| + | \end{align*} | ||
| + | |||
| + | ~sourodeepdeb | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=B|num-b=7|num-a=9}} | {{AMC12 box|year=2024|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 03:26, 14 November 2024
Problem
What value of 
satisfies
![\[\frac{\log_2x \cdot \log_3x}{\log_2x+\log_3x}=2?\]](http://latex.artofproblemsolving.com/a/d/6/ad6c3a949ac3db0de93e322df80e07a3a51e8a91.png)
Solution 1
We have
\begin{align*}
&\log_2x\cdot\log_3x=2(\log_2x+\log_3x) \\
&1=\frac{2(\log_2x+\log_3x)}{\log_2x\cdot\log_3x} \\
&1=2(\frac{1}{\log_3x}+\frac{1}{\log_2x}) \\
&1=2(\log_x3+\log_x2) \\
&\log_x6=\frac{1}{2} \\
&x^{\frac{1}{2}}=6 \\
&x=36
\end{align*}
so 
Solution 2 (Change of Base)
\begin{align*} \frac{\log_2x \cdot \log_3x}{\log_2x+\log_3x} &= 2 \\[6pt] \log_2x \cdot \log_3x &= 2(\log_2x+\log_3x) \\[6pt] \log_2x \cdot \log_3x &= 2\log_2x + 2\log_3x \\[6pt] \frac{\log x}{\log 2} \cdot \frac{\log x}{\log 3} &= 2\frac{\log x}{\log 2} + 2\frac{\log x}{\log 3} \\[6pt] \frac{(\log x)^2}{\log 2 \cdot \log 3} &= \frac{2\log x \cdot \log 3 + 2\log x \cdot \log 2}{\log 2 \cdot \log 3} \\[6pt] (\log x)^2 &= 2\log x \cdot \log 3 + 2\log x \cdot \log 2 \\[6pt] (\log x)^2 &= 2\log x(\log 2 + \log 3) \\[6pt] \log x &= 2(\log 2 + \log 3) \\[6pt] x &= 10^{2(\log 2 + \log 3)} \\[6pt] x &= (10^{\log 2} \cdot 10^{\log 3})^2 \\[6pt] x &= (2 \cdot 3)^2 = 6^2 = \boxed{\textbf{(C) }36} \end{align*}
~sourodeepdeb
See also
| 2024 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
