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Sloving exponential equations

  1. Feb 21, 2008 #1
    1. The problem statement, all variables and given/known data


    3. The attempt at a solution

    i have no clue as to how to solve this...if someone has a tutorial on another website of something on these type of questions...that would be helpful...can someone provide some insight as to how approach this problem?
  2. jcsd
  3. Feb 21, 2008 #2
    u can re-write as:

    0 = (2^x)^2 - 3* 2^X * 2^2 + 32
  4. Feb 21, 2008 #3

    How about now? Remember that, [tex]a^{x+y}=a^xa^y[/tex] Make a "substitution" if necessary.
  5. Feb 21, 2008 #4

    i see how it is done...i will try to do a number of these...will post if need more help.

    many thanks rocophysics and shawshank
  6. Feb 21, 2008 #5

    when i solve it....i end up with the answers x=0 or x=-1...the answer sheet says its just x=0....i did it multiple times...keep getting the same answer....which is right?
  7. Feb 21, 2008 #6



    [tex]2^x=4 \ \ \ 2^x=8[/tex]

    [tex]x=\frac{\ln 4}{\ln 2} \ \ \ x=\frac{\ln 8}{\ln 2}[/tex]

    Lol, omg! New problem, sorry. I was like wtf ... I'm not even getting 0 :p
  8. Feb 21, 2008 #7

    Simplifies to ...

    [tex]3^{x^2}=3^{\frac 1 x}[/tex]

    I'm getting 1 as my only answer.
    Last edited: Feb 21, 2008
  9. Feb 21, 2008 #8
    ohh crap...my bad...the answer i was looking at was for the pervious one.

    here is how i did it...tell me where i went wrong:







    x=0 or x=-1

    so i did something really wrong...did i violate a rule??
  10. Feb 21, 2008 #9
    [tex]27^x\cdot3^{x^2}=27^x\cdot3^{\frac 1 x}[/tex]

    [tex]3^{x^2}=3^{\frac 1 x}[/tex]

    [tex]x^2=\frac 1 x[/tex]

  11. Feb 21, 2008 #10
    ok...so i can't make the exponent 1/x into -x??
  12. Feb 21, 2008 #11
    You can, but why do it? Also, 27^x cancels nicely so it's just a cubic function with one real solution.
  13. Feb 22, 2008 #12


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    Homework Helper

    No, he cannot do it. 1/x cannot be change to -x.

    You seem to be confusing between the two:

    [tex]\frac{1}{a} = a ^ {-1}[/tex], this is a fine manipulation. :)

    However, [tex]\left( a ^ {\frac{1}{x}} = \sqrt[x]{a} \right) \neq \left( a ^ {-x} = \frac{1}{a ^ x} \right)[/tex] The two are completely different.

    You are wrong at the very first step.. Now, let's re-do it using ricophysics' way. :)
  14. Feb 22, 2008 #13
    Oh! Oops, I did not mean to mislead. I accidently read an exponent of -x, rather than a change to -x. Thanks VietDao!!! :-]
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