Difference between revisions of "2024 AIME I Problems/Problem 2"

Revision as of 12:42, 2 February 2024

There exist real numbers $x$ and $y$ , both greater than 1, such that $\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10$ . Find $xy$ .

Solution 1

By properties of logarithms, we can simplify the given equation to $x\log_xy=4y\log_yx=10$ . Let us break this into two separate equations: \begin{align*} x\log_xy&=10 \\ 4y\log_yx&=10. \\ \end{align*} We multiply the two equations to get: $4xy\left(\log_xy\log_yx\right)=100.$

Also by properties of logarithms, we know that $\log_ab\cdot\log_ba=1$ ; thus, $\log_xy\cdot\log_yx=1$ . Therefore, our equation simplifies to:

$4xy=100\implies xy=\boxed{025}.$

~Technodoggo

@@ Line 1: / Line 1: @@
-Determine all composite integers <math>n>1</math> that satisfy the following property: if <math>d_1,d_2,\dots,d_k</math> are all the positive divisors of <math>n</math> with <math>1=d_1<d_2<\dots<d_k=n</math>, then <math>d_i</math> divides <math>d_{i+1}+d_{i+2}</math> for every <math>1\le i \le k-2</math>.
+There exist real numbers <math>x</math> and <math>y</math>, both greater than 1, such that <math>\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10</math>. Find <math>xy</math>.
+==Solution 1==
+By properties of logarithms, we can simplify the given equation to <math>x\log_xy=4y\log_yx=10</math>. Let us break this into two separate equations:
+\begin{align*}
+x\log_xy&=10 \\
+y\log_yx&=10. \\
+\end{align*}
+We multiply the two equations to get:
+<cmath>4xy\left(\log_xy\log_yx\right)=100.</cmath>
+Also by properties of logarithms, we know that <math>\log_ab\cdot\log_ba=1</math>; thus, <math>\log_xy\cdot\log_yx=1</math>. Therefore, our equation simplifies to:
+<cmath>4xy=100\implies xy=\boxed{025}.</cmath>
+~Technodoggo

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Difference between revisions of "2024 AIME I Problems/Problem 2"

Revision as of 12:42, 2 February 2024

Solution 1