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Math Help - Solving Exponential Equations

  1. #1
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    Unhappy Solving Exponential Equations

    I need a quick explanation on how to solve 2^x = 30, and how to solve small equations like these when the unknown is an exponent?

    Thank you so much.
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  2. #2
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    Quote Originally Posted by needhelpplease1 View Post
    I need a quick explanation on how to solve 2^x = 30, and how to solve small equations like these when the unknown is an exponent?

    Thank you so much.
    use logs ...

    \ln(2^x) = \ln(30)

    x \cdot \ln(2) = \ln(30)

    x = \frac{\ln(30)}{\ln(2)}
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  3. #3
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    Quote Originally Posted by skeeter View Post
    use logs ...

    \ln(2^x) = \ln(30)

    x \cdot \ln(2) = \ln(30)

    x = \frac{\ln(30)}{\ln(2)}
    Sorry, but do you have time to explain one quick thing... did you use a calculator to do that, no right? you just knew to do 30/2 which means that x = 15 right?
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  4. #4
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    Quote Originally Posted by needhelpplease1 View Post
    Sorry, but do you have time to explain one quick thing... did you use a calculator to do that, no right? you just knew to do 30/2 which means that x = 15 right?
    no ... x \ne 15

    x = \frac{\ln(30)}{\ln(2)} \approx 4.907
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  5. #5
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    Quote Originally Posted by needhelpplease1 View Post
    I need a quick explanation on how to solve 2^x = 30, and how to solve small equations like these when the unknown is an exponent?

    Thank you so much.
    Consider that a^b = c \implies b = \log_ac

    Therefore 2^x = 30 \implies x = \log_230

    This will give you the same answer skeeter's solution provides.
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