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Math Help - Please help

  1. #1
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    Please help

    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
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by helpjoe View Post
    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
    \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 2^2 and 10 as 2*5:
    \left ( \frac{1}{5} \right ) ^m \times \left ( \frac{1}{2^2} \right )^{18} = \left ( \frac{1}{2(2*5)^{35}} \right )


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


    \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
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  3. #3
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    Quote Originally Posted by topsquark View Post
    \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 2^2 and 10 as 2*5:
    \left ( \frac{1}{5} \right ) ^m \times \left ( \frac{1}{2^2} \right )^{18} = \left ( \frac{1}{2(2*5)^{35}} \right )


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


    \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
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by helpjoe View Post
    thanks!!! btw, can you take the gmat for me
    Yes. I want $$. Lots and lots of $$.

    -Dan
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  5. #5
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    Hello, helpjoe!

    A variation on Ron's solution . . .


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

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

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

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

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

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

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