# Difference between revisions of "2005 AMC 12B Problems/Problem 13"

## Problem

Suppose that $4^{x_1}=5$, $5^{x_2}=6$, $6^{x_3}=7$, ... , $127^{x_{124}}=128$. What is $x_1x_2...x_{124}$?

$\mathrm{(A)}\ {{{2}}} \qquad \mathrm{(B)}\ {{{\frac{5}{2}}}} \qquad \mathrm{(C)}\ {{{3}}} \qquad \mathrm{(D)}\ {{{\frac{7}{2}}}} \qquad \mathrm{(E)}\ {{{4}}}$

## Solution

We see that we can re-write $4^{x_1}=5$, $5^{x_2}=6$, $6^{x_3}=7$, ... , $127^{x_{124}}=128$ as $\left(...\left(\left(\left(4^{x_1}\right)^{x_2}\right)^{x_3}\right)...\right)^{x_{124}}=128$ by using substitution. By using the properties of exponents, we know that $4^{x_1x_2...x_{124}}=128$.

$4^{x_1x_2...x_{124}}=128\\2^{2x_1x_2...x_{124}}=2^7\\2x_1x_2...x_{124}=7\\x_1x_2...x_{124}=\dfrac{7}{2}$

Therefore, the answer is $\boxed{\mathrm{(D)}\,\dfrac{7}{2}}$

## Alternate Solution

Changing $4^{x_1}=5$ to logarithmic form, we get ${x_1}=log_4 5$. We can rewrite this as ${x_1}=\dfrac{log_5}{log_4}$. Applying this to the rest, we get $x_1x_2...x_{124}=\dfrac{log5}{log4}*\dfrac{log6}{log5}*...*\dfrac{log128}{log127}=\dfrac{log5log6...log128}{log4log5...log127}=\dfrac{log128}{log4}=log_4128=log_4{2^7}=7*log_42=7*\dfrac{1}{2}=\boxed{\mathrm{(D)}\,\dfrac{7}{2}}$