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Math Help - Using Logarithms to solve for x (form a^x.b^y)

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    Using Logarithms to solve for x (form a^x.b^y)

    Hi there, if someone could please help me solve this problem I'd be most grateful:

    2^x-3 . 5^x+1 = 62.5
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    Quote Originally Posted by flashylightsmeow View Post
    Hi there, if someone could please help me solve this problem I'd be most grateful:

    2^x-3 . 5^x+1 = 62.5
    I am unsure of how your problem reads, but I think it is this...Please use () in the future.

    2^{x-3}\cdot 5^{x+1}=62.5 First multiply both sides by 2^4=16

    This gives
    2^{x+1}\cdot 5^{x+1}=1000 Now since the exponents on the left hand side are the same we can combine the bases to get

    10^{x+1}=1000 \iff 10^{x+1}=10^{3} \implies x+1=3 \implies x=2
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  3. #3
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    AH! Sorry about the layout error. So with problems like these its best to try and make the exponents the same?
    Thank you so much!!

    Jam
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