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 $x_1x_2...x_{124}=\dfrac{\log5}{\log4}\cdot\dfrac{\log6}{\log5}\cdot...\cdot\dfrac{\log128}{\log127}=\dfrac{\log5\cdot\log6\cdot...\cdot\log128}{\log4\cdot\log5\cdot...\cdot\log127}=\dfrac{\log128}{\log4}=\log_{4}{128}=\log_{4}{2^7}=7\cdot\log_{4}{2}=7\cdot\dfrac{1}{2}=\boxed{\mathrm{(D)}\,\dfrac{7}{2}}$

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2005 AMC 12B Problems/Problem 13

Contents

Problem

Solution

Alternate Solution

See also