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Math Help - Solving for x

  1. #1
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    Solving for x

    Can anyone help me out on this equation, i am doing exponents and logarithms, and can't seem to understand this, i get that you are supposed to make the base the same and then solve for x using the exponents, but how do you make the base the same in this function.

    7 ^3x+1=5^x

    The 3x+1 is actually the exponent
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by nuckers View Post
    Can anyone help me out on this equation, i am doing exponents and logarithms, and can't seem to understand this, i get that you are supposed to make the base the same and then solve for x using the exponents, but how do you make the base the same in this function.

    7 ^3x+1=5^x

    The 3x+1 is actually the exponent
    (3x+1)\ln(7) = x\ln(5)

    You can distribute on the LHS as with any normal algebra
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by nuckers View Post
    Can anyone help me out on this equation, i am doing exponents and logarithms, and can't seem to understand this, i get that you are supposed to make the base the same and then solve for x using the exponents, but how do you make the base the same in this function.

    7 ^3x+1=5^x

    The 3x+1 is actually the exponent
    7^{3x+1} = 5^x

    log(7^{3x+1}) = log(5^x)

    (3x+1) \mbox{log(7)} = x \mbox{log(5)}

    \frac{3x+1}{x}= \frac{\mbox{log(5)}}{\mbox{log(7)}}

    3+ \frac{1}{x} = \frac{\mbox{log(5)}}{\mbox{log(7)}}

    finish it..
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  4. #4
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    Quote Originally Posted by harish21 View Post
    7^{3x+1} = 5^x

    log(7^{3x+1}) = log(5^x)

    (3x+1) \mbox{log(7)} = x \mbox{log(5)}

    \frac{3x+1}{x}= \frac{\mbox{log(5)}}{\mbox{log(7)}}

    3+ \frac{1}{x} = \frac{\mbox{log(5)}}{\mbox{log(7)}}

    finish it..
    So i get -.46021234 for my final answer, does that sound right?
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by nuckers View Post
    So i get -.46021234 for my final answer, does that sound right?
    That's correct ^_^

    You can test it by putting it into your original equation. If the two sides are equal than the answer is fine
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