• Oct 20th 2006, 12:54 PM
helpjoe
here is the question:

(1/5) to the m'th power multiplied by (1/4) to the 18th power= 1/(2(10) to the 35th power)

the last phrase looks a little confusing so let me try to explain. it is 1 over 2 times ten, which ten is to the 35th power.

the first part is 1 divided by five, and the total expression to the m'th power and the second part is 1 divided by 4, all to the 18th power.

any help would be greatly appreciated and please show work!!! i got the gmat exam tomorrow, saturday!!! if anyone has an understanding of this problem and could call me, i would greatly appreciate it. my number is 954-465-5395. thanks in advance
• Oct 20th 2006, 01:13 PM
topsquark
Quote:

Originally Posted by helpjoe
here is the question:

(1/5) to the m'th power multiplied by (1/4) to the 18th power= 1/(2(10) to the 35th power)

the last phrase looks a little confusing so let me try to explain. it is 1 over 2 times ten, which ten is to the 35th power.

the first part is 1 divided by five, and the total expression to the m'th power and the second part is 1 divided by 4, all to the 18th power.

any help would be greatly appreciated and please show work!!! i got the gmat exam tomorrow, saturday!!! if anyone has an understanding of this problem and could call me, i would greatly appreciate it. my number is 954-465-5395. thanks in advance

$\displaystyle \left ( \frac{1}{5} \right ) ^m \times \left ( \frac{1}{4} \right )^{18} = \left ( \frac{1}{2(10)^{35}} \right )$

I presume you want to find "m"?

First express 4 as $\displaystyle 2^2$ and 10 as 2*5:
$\displaystyle \left ( \frac{1}{5} \right ) ^m \times \left ( \frac{1}{2^2} \right )^{18} = \left ( \frac{1}{2(2*5)^{35}} \right )$

$\displaystyle \left ( \frac{1}{5} \right ) ^m \times \left ( \frac{1}{2} \right )^{36} = \left ( \frac{1}{2(2)^{35}(5)^{35}} \right )$

$\displaystyle \left ( \frac{1}{5} \right ) ^m \times \left ( \frac{1}{2} \right )^{36} = \left ( \frac{1}{2^{36}5^{35}} \right )$

By inspection m = 35.

-Dan
• Oct 20th 2006, 01:29 PM
helpjoe
Quote:

Originally Posted by topsquark
$\displaystyle \left ( \frac{1}{5} \right ) ^m \times \left ( \frac{1}{4} \right )^{18} = \left ( \frac{1}{2(10)^{35}} \right )$

I presume you want to find "m"?

First express 4 as $\displaystyle 2^2$ and 10 as 2*5:
$\displaystyle \left ( \frac{1}{5} \right ) ^m \times \left ( \frac{1}{2^2} \right )^{18} = \left ( \frac{1}{2(2*5)^{35}} \right )$

$\displaystyle \left ( \frac{1}{5} \right ) ^m \times \left ( \frac{1}{2} \right )^{36} = \left ( \frac{1}{2(2)^{35}(5)^{35}} \right )$

$\displaystyle \left ( \frac{1}{5} \right ) ^m \times \left ( \frac{1}{2} \right )^{36} = \left ( \frac{1}{2^{36}5^{35}} \right )$

By inspection m = 35.

-Dan

thanks!!! btw, can you take the gmat for me :)
• Oct 20th 2006, 01:47 PM
topsquark
Quote:

Originally Posted by helpjoe
thanks!!! btw, can you take the gmat for me :)

Yes. I want $$. Lots and lots of$$. :)

-Dan
• Oct 20th 2006, 04:13 PM
Soroban
Hello, helpjoe!

A variation on Ron's solution . . .

Quote:

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

We have: .$\displaystyle \frac{1}{5^m}\cdot\frac{1}{4^{18}} \:=\:\frac{1}{2\cdot10^{35}}$

Take reciprocals: .$\displaystyle 5^m\cdot4^{18} \:=\:2\cdot10^{35}$

Then we have: .$\displaystyle 5^m(2^2)^{18} \:=\:2(2\cdot5)^{35} \:=\:2\cdot2^{35}\cdot5^{35}$

. . . . . . . . . . . .$\displaystyle 5^m \cdot 2^{36} \:=\: 2^{36} \cdot 5^{35}$

Therefore: .$\displaystyle 5^m\:=\:5^{35}\quad\Rightarrow\quad m \,= \,35$