# GMAT question

• March 26th 2011, 10:43 AM
integrity
GMAT question
I put this one under the high school level because it really shouldn't be that hard, but it has me stumped. Any advice?

Question: what is the value of m in the following expression?

(1/5)^m * (1/4)^18 = 1/(2*(10^35))

The denominator exponent bases on the LHS are 5 and 4, so they can't be combined. I was thinking about 4 = 2^2, but that isn't helping me either.

Anything thoughts are much appreciated.
--integrity
• March 26th 2011, 10:48 AM
mathfun
${\left( {\tfrac{1}{5}} \right)^m} \cdot {\left( {\tfrac{1}{4}} \right)^{18}} = \tfrac{1}{2} \cdot {10^{ - 35}} \Leftrightarrow {\left( {\tfrac{1}{5}} \right)^m} = \tfrac{1}{2}{\left( {\tfrac{1}{2}} \right)^{ - 36}}{10^{ - 35}} \Leftrightarrow {\left( {\tfrac{1}{5}} \right)^m} = {\left( {\tfrac{1}{5}} \right)^{35}} \Leftrightarrow m = 35$
• March 26th 2011, 11:09 AM
integrity
Thanks for your quick response mathfun. Much appreciated!
I went through your calculations carefully and I think a negative sign was dropped in the 4th step, which would mean the correct answer is ultimately 35. I think.
• March 26th 2011, 01:48 PM
Soroban
Hello, integrity!

Quote:

$\displaystyle \text{Solve for }m:\;\left(\frac{1}{5}\right)^m\left(\frac{1}{4}\r ight)^{18} \:=\:\frac{1}{2\cdot10^{35}}$

$\text{I did it like this . . .}$

$\displaystyle\text{We have: }\:\frac{1}{5^m}\cdot\frac{1}{(2^2)^{18}} \;=\;\frac{1}{2\cdot(2\cdot5)^{35}}$

. . . . . . . . . . $\displaystyle \frac{1}{5^m}\cdot\frac{1}{2^{36}} \;=\;\frac{1}{2\cdot2^{35}\cdot5^{35}}$

. . . . . . . . . . $\displaystyle \frac{1}{5^m\cdot2^{36}} \;=\;\frac{1}{2^{36}\cdot5^{35}}$

$\displaystyle \text{Multiply by }2^{36}\!:\;\;\frac{1}{5^m} \;=\;\frac{1}{5^{35}}$

. . . . . . . . . . . . . $\displaystyle 5^{-m} \;=\;5^{-35}$

. . . . . . . . . . . . . $-m \;=\;-35$

. . . . . . . . . . . . . . $m \;=\;35$

• March 26th 2011, 01:52 PM
integrity
That's it! Thanks very much.