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Some Confusion with an Exponential Equation

  1. May 7, 2012 #1

    M83

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    I'm reviewing for my final exam so one of the practice problems is:

    5= 3^(x+5)

    Here's my attempt at it:

    ln 5= x+5 ln 3

    ln 5 / ln 3 = x+5
    (ln 5 / ln 3)-5= x
    1.46-5 ≈ x
    -3.54 ≈ x

    I checked my answer and I get 3^1.46 ≈ 4.97 so rounding it up gives me 5 since I rounded off 1.46. However when I plug this equation into Mathway it gives me an answer of -3.39. I tried figuring out why (I know if you pay a fee you can view their steps, but I don't have the money at the moment) and eventually came to the conclusion that this is how they did it:

    5= 3^(x+5)

    x+5 = ln 5
    x = ln 5 - 5
    x ≈ -3.39

    But when I check their answer I get 3^1.61 ≈ 5.86

    I think Mathway is wrong, but maybe I'm missing something. Which method is the correct one for solving this equation? Thanks for any help I receive.
     
  2. jcsd
  3. May 7, 2012 #2
    Don't worry, your answer is correct.
    Also, give Wolfram|Alpha a try; it's free and it shows steps (although they're not always very helpful depending on the problem).
     
  4. May 7, 2012 #3

    M83

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    Thanks Bohrok. I'll give Wolfram a try.
     
  5. May 7, 2012 #4

    Mark44

    Staff: Mentor

    The equation above needs parentheses.

    What you meant was
    ln 5= (x+5) ln 3

    Because of the higher precedence of multiplication over addition, what you wrote would be interpreted as
    ln 5= x+ (5 ln 3)
     
  6. May 7, 2012 #5

    M83

    User Avatar

    Thanks for the correction.
     
  7. May 7, 2012 #6
    5 = 3^x * 3^5

    5/(3^5) = 3^x

    x = log_3(5/(3^5)) whatever that turns out to be. Let's go further ...

    x = log_3(5/(3^5)) = log_3(5) - log_3(3^5)

    = log_3(5) - 5

    = -3.53502647928207283280295959232135960369206763333395...

    that last step from Wolfram Alpha. So I think what you had initially is correct.
     
  8. May 12, 2012 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You could also take the logarithm directly:
    [tex]5= 3^{x+ 5}[/tex]
    [tex]log(5)= log(3^{x+5})= (x+ 5)log(3)[/tex]
    so that
    [tex]x+ 5= \frac{log(5)}{log(3)}[/tex]
    and then
    [tex]x= \frac{log(5)}{log(3)}- 5[/tex].

    Are you required to write a decimal answer (which can only be approximate)? I would leave the answer as the above fraction.
     
  9. May 15, 2012 #8
    your answer is correct, with these questions always take the ln/e and solve from their, quite straight forward :)
     
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