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Thread: Exponential Equations with Logarithms

  1. #1
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    Exponential Equations with Logarithms

    Hi everyone,
    I need some help with these exponential equations. I am required to use logarithms to solve them. I have hit a wall!

    1. 2^x=30
    2. 5^x-1=3^x
    3. 3.5^3x+1=65.4
    4. 16^x-4=3^3-x
    5. 7^x-2=5^x


    Help/explanations would be greatly appreciated!
    Thanks so much!
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  2. #2
    Senior Member eumyang's Avatar
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    Please learn LaTeX, or at least use grouping symbols. For instance, in #2 is it supposed to be
    $\displaystyle 5^x - 1 = 3^x$ or $\displaystyle 5^{x-1} = 3^x$?

    I'll do the first, as it doesn't look ambiguous. To isolate the x, take the log base-2 of both sides:
    $\displaystyle \begin{aligned}
    2^x &= 30 \\
    \log_2 2^x &= \log_2 30 \\
    x &= \log_2 30
    \end{aligned}$

    If you need to state your answer as a decimal, use the change of base formula to change into either natural logs (base e) or common logs (base 10), and then plug into the calculator:
    $\displaystyle \log_2 30 = \dfrac{\log 30}{\log 2} \approx 4.907$
    OR
    $\displaystyle \log_2 30 = \dfrac{\ln 30}{\ln 2} \approx 4.907$
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by allegroconbrio95 View Post
    Hi everyone,
    I need some help with these exponential equations. I am required to use logarithms to solve them. I have hit a wall!

    1. 2^x=30
    2. 5^x-1=3^x
    3. 3.5^3x+1=65.4
    4. 16^x-4=3^3-x
    5. 7^x-2=5^x


    Help/explanations would be greatly appreciated!
    Thanks so much!
    Hi allegroconbrio95,

    [1] Using natural log:

    $\displaystyle 2^x=30$

    $\displaystyle \ln 2^x= \ln 30$

    $\displaystyle x\ln 2=\ln 30$

    $\displaystyle x=\frac{\ln 30}{\ln 2}$

    $\displaystyle x \approx 4.9069$



    [2] $\displaystyle 5^{x-1}=3^x$

    $\displaystyle \ln 5^{x-1}=\ln 3^x$

    $\displaystyle (x-1)\ln 5 = x \ln 3$

    $\displaystyle x \ln 5-\ln 5=x \ln 3$

    $\displaystyle x \ln 5 - x \ln 3 = \ln 5$

    $\displaystyle x(\ln 5 - \ln 3)= \ln 5$

    $\displaystyle x=\frac{\ln 5}{\ln 5-\ln 3}$

    $\displaystyle x \approx 3.1507$

    Now, the rest are similar to [2] so you have a pretty good model to use. Good luck.



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