Difference between revisions of "2020 AIME II Problems/Problem 3"

Revision as of 10:55, 14 September 2020

Problem

The value of $x$ that satisfies $\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

Let $\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n$ . Based on the equation, we get $(2^x)^n=3^{20}$ and $(2^{x+3})^n=3^{2020}$ . Expanding the second equation, we get $8^n\cdot2^{xn}=3^{2020}$ . Substituting the first equation in, we get $8^n\cdot3^{20}=3^{2020}$ , so $8^n=3^{2000}$ . Taking the 100th root, we get $8^{\frac{n}{100}}=3^{20}$ . Therefore, $(2^{\frac{3}{100}})^n=3^{20}$ , and using the our first equation( $2^{xn}=3^{20}$ ), we get $x=\frac{3}{100}$ and the answer is $\boxed{103}$ . ~rayfish

Easiest Solution

Recall the identity $\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b$ (which is easily proven using exponents or change of base). Then this problem turns into $\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3$ Divide $\log_{2} 3$ from both sides. And we are left with $\frac{20}{x}=\frac{2020}{x+3}$ .Solving this simple equation we get $x = \tfrac{3}{100} \Rightarrow \boxed{103}$ ~mlgjeffdoge21

Solution 2

Because $\log_a{b^c}=c\log_a{b},$ we have that $20\log_{2^x} 3 = 2020\log_{2^{x+3}} 3,$ or $\log_{2^x} 3 = 101\log_{2^{x+3}} 3.$ Since $\log_a{b}=\dfrac{1}{\log_b{a}},$ $\log_{2^x} 3=\dfrac{1}{\log_{3} 2^x},$ and $101\log_{2^{x+3}} 3=101\dfrac{1}{\log_{3}2^{x+3}},$ thus resulting in $\log_{3}2^{x+3}=101\log_{3} 2^x,$ or $\log_{3}2^{x+3}=\log_{3} 2^{101x}.$ We remove the base 3 logarithm and the power of 2 to yield $x+3=101x,$ or $x=\dfrac{3}{100}.$

Our answer is $\boxed{3+100=103}.$ ~ OreoChocolate

Solution 3 (Official MAA)

Using the Change of Base Formula to convert the logarithms in the given equation to base $2$ yields $\frac{\log_2 3^{20}}{\log_2 2^x} = \frac{\log_2 3^{2020}}{\log_2 2^{x+3}}, \text{~ and then ~} \frac{20\log_2 3}{x\cdot\log_2 2} = \frac{2020\log_2 3}{(x+3)\log_2 2}.$ Canceling the logarithm factors then yields $\frac{20}x = \frac{2020}{x+3},$ which has solution $x = \frac3{100}.$ The requested sum is $3 + 100 = 103$ .

Solution 4

$\log_{2^x} 3^{20} = 2^{xy} = 3^{20}$

$\log_{2^{x+3}} 3^{2020} = (2^{x+3})^y = 3^{2020}$

$3^{2020} = (3^{20})^{101}$

$(2^{xy})^{101} = (2^{x+3})^y = 3^{2020}$

$(2^{xy})^{101} = (2^{x+3})^y = 2^{101xy} = 2^{xy+3y}$ $\Rightarrow$ $101xy = xy + 3y$ $\Rightarrow$ $101xy = y(x+3)$

$101x = x + 3$

$100x = 3$

$x = \frac{3}{100}$

$100 + 3 =$ \boxed{103}$ ~Airplanes2007

Video Solution

https://youtu.be/lPr4fYEoXi0 ~ CNCM

Video Solution 2

https://www.youtube.com/watch?v=x0QznvXcwHY?t=528

~IceMatrix

Video Solution 3

https://youtu.be/-CkEF5nWOaI

~avn

Video Solution 4

https://www.youtube.com/watch?v=2TSNY2DDUbQ&t=3s ~ MathEx

@@ Line 26: / Line 26: @@
 ==Solution 4==
 <math>\log_{2^x} 3^{20} = 2^{xy} = 3^{20}</math>
 <math>\log_{2^{x+3}} 3^{2020} = (2^{x+3})^y = 3^{2020}</math>
 <math>3^{2020} = (3^{20})^{101}</math>
 <math>(2^{xy})^{101} = (2^{x+3})^y = 3^{2020}</math>
-<math>(2^{xy})^{101} = (2^{x+3})^y = 2^{101xy} = 2^{xy+3y}</math> \Rightarrow <math>101xy = xy + 3y</math> \Rightarrow <math>101xy = y(x+3)</math>
+<math>(2^{xy})^{101} = (2^{x+3})^y = 2^{101xy} = 2^{xy+3y}</math> <math>\Rightarrow</math> <math>101xy = xy + 3y</math> <math>\Rightarrow</math> <math>101xy = y(x+3)</math>
 <math>101x = x + 3</math>
 <math>100x = 3</math>
 <math>x = \frac{3}{100}</math>
 <math>100 + 3 = </math>\boxed{103}$ ~Airplanes2007

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