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Math Help - Solving for Exponent (GMAT prep)

  1. #1
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    Solving for Exponent (GMAT prep)

    Not sure if this is the right forum for this. I'm new here

    I'm studying to take the GMAT and I'm stumped on the following problem:

    5^21 * 4^11 = 2 * 10^n

    I need to solve for n. The answer I believe is 21, but I don't know how to arrive at the answer without a calculator. So I'm asking how to solve this without actually calculating the 5^21 and 4^11.

    All the review material I've been reading talks about having like bases before combining exponents and I'm not seeing how to get like bases in this problem.
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  2. #2
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    Quote Originally Posted by frostbyteying View Post
    Not sure if this is the right forum for this. I'm new here

    I'm studying to take the GMAT and I'm stumped on the following problem:

    5^21 * 4^11 = 2 * 10^n

    I need to solve for n. The answer I believe is 21, but I don't know how to arrive at the answer without a calculator. So I'm asking how to solve this without actually calculating the 5^21 and 4^11.

    All the review material I've been reading talks about having like bases before combining exponents and I'm not seeing how to get like bases in this problem.
    Write out both sides of this in terms of prime fractors:

    5^{21}\ 2^{22}=5^n\ 2^{n+1}

    and as prime factorisation is unique, we have that n=21

    RonL
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  3. #3
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    Thanks for the response!

    I see how you get n=21 from the equation you've written and I understand how you got the left side. I'm confused as to how you got the right side of the equation though.

    How does 2 * 10^n become 5^n * 2^(n+1) ?

    Can you walk me through the transition?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by frostbyteying View Post
    Thanks for the response!

    I see how you get n=21 from the equation you've written and I understand how you got the left side. I'm confused as to how you got the right side of the equation though.

    How does 2 * 10^n become 5^n * 2^(n+1) ?

    Can you walk me through the transition?
    First break the 10 into its prime factors 10=2 \times 5, then 10^n=(2\times 5)^n=2^n\times 5^n, so we have:

    <br />
2 \times 10^n= 2\times 2^n \times 5^n=2^{n+1}\times 5^n<br />

    RonL
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  5. #5
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    Thanks a lot for your help CaptainBlack! That was just the explanation I needed. Everything makes perfect sense.
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