Difference between revisions of "2005 AMC 12B Problems/Problem 13"
m |
|||
| Line 1: | Line 1: | ||
| − | |||
== Problem == | == Problem == | ||
| − | {{ | + | Suppose that <math>4^{x_1}=5</math>, <math>5^{x_2}=6</math>, <math>6^{x_3}=7</math>, ... , <math>127^{x_{124}}=128</math>. What is <math>x_1x_2...x_{124}</math>? |
| + | |||
| + | <math>\mathrm{(A)}\ {{{2}}} \qquad \mathrm{(B)}\ {{{\frac{5}{2}}}} \qquad \mathrm{(C)}\ {{{3}}} \qquad \mathrm{(D)}\ {{{\frac{7}{2}}}} \qquad \mathrm{(E)}\ {{{4}}}</math> | ||
== Solution == | == Solution == | ||
| + | We see that we can re-write <math>4^{x_1}=5</math>, <math>5^{x_2}=6</math>, <math>6^{x_3}=7</math>, ... , <math>127^{x_{124}}=128</math> as <math>\left(...\left(\left(\left(4^{x_1}\right)^{x_2}\right)^{x_3}\right)...\right)^{x_{124}}=128</math> by using substitution. By using the properties of exponents, we know that <math>4^{x_1x_2...x_{124}}=128</math>. | ||
| + | |||
| + | <math>\begin{align*}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}\end{align*}</math> | ||
| + | Therefore, the answer is <math>\boxed{\mathrm{(D)}\,\dfrac{7}{2}}</math> | ||
== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2005|ab=B|num-b=12|num-a=14}} | |
Revision as of 12:53, 1 July 2011
Problem
Suppose that 
, 
, 
, ... , 
. What is 
?
Solution
We see that we can re-write , , , ... , as 
by using substitution. By using the properties of exponents, we know that 
.
$\begin{align*}4^{x_1x_2...x_{124}}=128\\2^{2x_1x_2...x_{124}}=2^7\\2x_1x_2...x_{124}=7\\x_1x_2...x_{124}=\dfrac{7}{2}\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Therefore, the answer is 
See also
| 2005 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
