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Math Help - [SOLVED] Algebra indices simultaneous equation

  1. #1
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    [SOLVED] Algebra indices simultaneous equation

    Solve the simultaneous equation x-y=1 and (2^x)(3^y)=432

    Do you make x=1+y first and then put it into the other equation?

    I don't know what to do once I've substituted or if that's what you have to do.

    Thank you in advance for any help
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  2. #2
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    Quote Originally Posted by greghunter View Post
    Solve the simultaneous equation x-y=1 and (2^x)(3^y)=432

    Do you make x=1+y first and then put it into the other equation?

    I don't know what to do once I've substituted or if that's what you have to do.

    Thank you in advance for any help
    Yes you're on the right track.

    So you should end up with 2^{1 + y}\cdot 3^y = 432

    2^1 \cdot 2^y \cdot 3^y = 432

    2\cdot\left(2\cdot 3\right)^y = 432

    2\cdot 6^y = 432

    6^y = 216

    6^y = 6^3

    y = 3.


    Since x = 1 + y we find that x = 4.
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  3. #3
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    Quote Originally Posted by greghunter View Post
    Solve the simultaneous equation x-y=1 and (2^x)(3^y)=432

    Do you make x=1+y first and then put it into the other equation?

    I don't know what to do once I've substituted or if that's what you have to do.

    Thank you in advance for any help
    1. x-y=1~\implies~y = x-1

    2. (2^x)(3^y)=432~\implies~2^x\cdot 3^{x-1}=432

    2^x\cdot 3^x=1296 = 6^4

    6^x = 6^4~\implies~ x= 4~\wedge~y=3
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  4. #4
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    Cheers guys!
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