Difference between revisions of "2024 AMC 12B Problems/Problem 8"
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Then, from our definitions of <math>a</math> and <math>b</math>, <math>2^a=x</math> and <math>3^b=x</math>, so <math>2^a=3^b.</math> Taking the logarithm base <math>3</math> of both sides of this equation gives us <math>\log_3 2^a=b</math>, hence <math>a \log_3 2=b.</math> | Then, from our definitions of <math>a</math> and <math>b</math>, <math>2^a=x</math> and <math>3^b=x</math>, so <math>2^a=3^b.</math> Taking the logarithm base <math>3</math> of both sides of this equation gives us <math>\log_3 2^a=b</math>, hence <math>a \log_3 2=b.</math> | ||
Now, we substitute <math>a \log_3 2</math> for <math>b</math> in the equation, which gives <cmath>\frac{a \cdot a \log_3 2}{a+a \log_3 2}=2.</cmath>Notice that we can factor out an <math>a</math> in the numerator and denominator, if <math>a \neq 0,</math> and doing so yields <cmath>\frac{a \log_3 2}{1+\log_3 2}=2.</cmath> We know that <math>1= \log_3 3,</math> so putting that in gives us <cmath>\frac{a \log_3 2}{\log_3 3+\log_3 2}=2 \implies \frac{a \log_3 2}{\log_3 6}=2.</cmath>So, <math>a=2 \cdot \frac{\log_3 6}{\log_3 2}</math>, which, using the change of base formula in reverse, is equivalent to <math>2 \cdot \log_2 6,</math> thus, <cmath>a= \log _2 6^2= \log _2 36.</cmath> Finally, using our original definition of <math>a,</math> we have <cmath>a = \log_2 x=\log_2 36,</cmath> so <math>x=\boxed{\textbf{(C) }36}.</math> | Now, we substitute <math>a \log_3 2</math> for <math>b</math> in the equation, which gives <cmath>\frac{a \cdot a \log_3 2}{a+a \log_3 2}=2.</cmath>Notice that we can factor out an <math>a</math> in the numerator and denominator, if <math>a \neq 0,</math> and doing so yields <cmath>\frac{a \log_3 2}{1+\log_3 2}=2.</cmath> We know that <math>1= \log_3 3,</math> so putting that in gives us <cmath>\frac{a \log_3 2}{\log_3 3+\log_3 2}=2 \implies \frac{a \log_3 2}{\log_3 6}=2.</cmath>So, <math>a=2 \cdot \frac{\log_3 6}{\log_3 2}</math>, which, using the change of base formula in reverse, is equivalent to <math>2 \cdot \log_2 6,</math> thus, <cmath>a= \log _2 6^2= \log _2 36.</cmath> Finally, using our original definition of <math>a,</math> we have <cmath>a = \log_2 x=\log_2 36,</cmath> so <math>x=\boxed{\textbf{(C) }36}.</math> | ||
| + | |||
| + | ~hdanger | ||
==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 12:42, 18 November 2024
Contents
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 
~kafuu_chino
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
Solution 3 (Using Variables)
Let 
and 
. This gives us the equation ![\[\frac{ab}{a+b}=2\]](http://latex.artofproblemsolving.com/9/e/5/9e56dc2413fabb288ec34d33c82083235c9f2e19.png)
Then, from our definitions of 
and 
, 
and 
, so 
Taking the logarithm base 
of both sides of this equation gives us 
, hence 
Now, we substitute 
for in the equation, which gives ![\[\frac{a \cdot a \log_3 2}{a+a \log_3 2}=2.\]](http://latex.artofproblemsolving.com/e/c/3/ec33238b44717c8eda464ac0502d5cd16b799cfe.png)
Notice that we can factor out an in the numerator and denominator, if 
and doing so yields ![\[\frac{a \log_3 2}{1+\log_3 2}=2.\]](http://latex.artofproblemsolving.com/f/c/d/fcdbc3011a56e65d1d324b4168e51354fd2cfae9.png)
We know that 
so putting that in gives us ![\[\frac{a \log_3 2}{\log_3 3+\log_3 2}=2 \implies \frac{a \log_3 2}{\log_3 6}=2.\]](http://latex.artofproblemsolving.com/1/0/a/10a882e7115ea41c6144a141b0e764c10248bdf4.png)
So, 
, which, using the change of base formula in reverse, is equivalent to 
thus, ![\[a= \log _2 6^2= \log _2 36.\]](http://latex.artofproblemsolving.com/f/4/b/f4bf367095afe41ca6f02031123196b408f70415.png)
Finally, using our original definition of 
we have ![\[a = \log_2 x=\log_2 36,\]](http://latex.artofproblemsolving.com/5/2/0/520b511b3b52198d070df61c7e14ba37683b85e4.png)
so 
~hdanger
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. 
