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Math Help - Solving for X in both sides

  1. #1
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    Solving for X in both sides

    3^(7x-1) = 5^(3x+2)

    I've tried to take the natural log of both sides, and pull x out of the exponent, but I can't seem to get the right answer.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Velvet Love View Post
    3^(7x-1) = 5^(3x+2)

    I've tried to take the natural log of both sides, and pull x out of the exponent, but I can't seem to get the right answer.
    \ln{3^{(7x-1)}}=\ln{5^{(3x+2)}}

    \Rightarrow(7x-1)\ln{3}=(3x+2)\ln{5}

    Can you take it from here?
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  3. #3
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    That's what I've gotten it down to. Then I end up with ((ln(5)/ln(3))+3)/4, which is wrong. I'm rusty at pre-cal.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Velvet Love View Post
    That's what I've gotten it down to. Then I end up with ((ln(5)/ln(3))+3)/4, which is wrong. I'm rusty at pre-cal.
    Distribute:

    (7\ln3)x-\ln{3}=(3\ln{5})x+2\ln{5}

    Isolate x:

    (7\ln3)x-(3\ln{5})x=2\ln{5}+\ln3

    x(7\ln{3}-3\ln{5})=2\ln{5}+\ln3

    And now divide...
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  5. #5
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    Thank you alot. That helped a lot. I'm glad I learn best by example
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  6. #6
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    I worked out 1.51..... is that correct?
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  7. #7
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    Quote Originally Posted by sanddollar View Post
    I worked out 1.51..... is that correct?
    yes, if your answer is 3s.f.
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  8. #8
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    Talking

    Quote Originally Posted by Velvet Love View Post
    I end up with ((ln(5)/ln(3))+3)/4, which is wrong.
    It'll be a lot easier to figure out where you might be going wrong if you show your steps.
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