2005 AMC 12B Problems/Problem 13

Revision as of 13:35, 20 June 2015 by Jayjayliu (talk | contribs) (Solution)

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 $\boxed{\mathrm{(D)}\,\dfrac{7}{2}}$

See also

2005 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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. AMC logo.png

Invalid username
Login to AoPS