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Math Help - problem with logs and exponents

  1. #1
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    problem with logs and exponents

    I have the following formula I am struggling with.

    (3.4)^2x+3=8.5

    This is what I have so far:

    ^2x+3=8.5/3.4

    (2x+3)log2.5

    0.7958x and 1.937

    To be honest not to sure if this is right or if it is where to go from here.

    All help much appreciated.
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  2. #2
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    Re: problem with logs and exponents

    It is not clear what is equality you are dealing with. Is that :

    (3.4)^{2x+3}=8.5 or (3.4)^{2x}+3=8.5 or something else...
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  3. #3
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    Re: problem with logs and exponents

    it's the first one.
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  4. #4
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    Re: problem with logs and exponents

    Your method is presented terribly - I don't understand what you've tried at all. Take logs of both sides to begin with, and then use the rule that \log(x^n)=n\log(x) and then isolate the x term as normal.
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  5. #5
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    Re: problem with logs and exponents

    Quote Originally Posted by chrisn30 View Post
    I have the following formula I am struggling with.

    (3.4)^2x+3=8.5

    This is what I have so far:

    ^2x+3=8.5/3.4

    (2x+3)log2.5

    0.7958x and 1.937

    To be honest not to sure if this is right or if it is where to go from here.

    All help much appreciated.
    You can't divide the base and leave the exponent "floating". 3.4^{2x+3} is 3.4 multiplied by itself 2x+3 times so if you did divide by 3.4 you'd have it multiplied by itself 2x+2 times which gets you nowhere.

    Instead use the addition laws of exponents: 3.4^{2x+3} = 3.4^{2x} \times \underbrace{3.4^3}_{\text{this is a number}}


    We can divide both sides by 3.4^3 since it's a constant:

    3.4^{2x} = \dfrac{8.5}{3.4^3}

    You can now take logs using your favourite base.
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  6. #6
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    Re: problem with logs and exponents

    In that case :

    2x+3=\log_{3.4} 8.5 \Rightarrow x= \frac{\log_{3.4} ({8.5})-3}{2}
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  7. #7
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    e^(i*pi)'s Avatar
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    Re: problem with logs and exponents

    Alternatively, using base e:

    2x \ln(3.4) = \ln \left(\dfrac{8.5}{3.4^3}\right)

    x = \dfrac{\ln \left(\dfrac{8.5}{3.4^3}\right)}{2\ln(3.4)} = \dfrac{\ln(8.5) - 3\ln(3.4)}{2\ln(3.4)} = \dfrac{\ln(8.5)}{2\ln(3.4)} - \dfrac{3}{2}

    This is the same as princeps' answer which can be verified using the change of base rule
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