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Find x for x^(log 5)+5^(log x)=50

  1. May 28, 2014 #1
    1. The problem statement, all variables and given/known data

    Find ##x## if ##x^{log 5}+5^{log x}=50##

    2. Relevant equations



    3. The attempt at a solution

    I tried ##x^{log 5}=50-5^{log x}##

    and vice versa. Then substituting the value of x in another equation, I end up with irritating results like
    ##50=50, 0=0, 1=1, x^{log 5}=x^{log 5} ## etc.

    Can anyone give me a hint?
     
  2. jcsd
  3. May 28, 2014 #2

    ehild

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    Use the identity ##a= e^{\log(a)}##

    ehild
     
  4. May 28, 2014 #3

    Mentallic

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    When you end up with a result like that it means that your substitution was a solution to the equality. If you end up with 1=0 or any other variation then your substitution wasn't a solution. The problem you have however is that this hasn't gotten you anywhere closer to finding what x on its own is.

    Do what ehild suggested to try solve for x, and then when you have x=... (where it's a number on the right side and hence has no x value in it), try plugging that value in to your equation again and see if you can get it down to 1=1 or 0=0 etc. If you do, you've found a correct value of x.
     
  5. May 28, 2014 #4

    Ray Vickson

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    Write ##x = e^{\, \log \, x}, \; 5 = e^{\, \log \, 5}##.
     
  6. May 28, 2014 #5
    Thanks I got it ##x=50## also ##x^{\log 5}=5^{\log x}##.
     
  7. May 28, 2014 #6

    ehild

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    What is the base of the logarithm?

    ehild
     
  8. May 28, 2014 #7

    CAF123

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    Do you mean to say your answer is that x = 50?
     
  9. May 28, 2014 #8

    adjacent

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    That's wrong.##x \neq 50## Try substituting it into the equation and see if it shows correct
     
  10. May 28, 2014 #9
    ##\\x^{\log 5}+5^{\log x}=50\\\\##

    [itex]\left(10^{\log x }\right )^{\log 5}+\left(10^{\log 5 }\right )^{\log x} =50\\\\[/itex]

    [itex]10^{\log x \log 5}+10^{\log 5 \log x}=50[/itex]

    ##\text{Since both are same, } x^{log 5}=5^{log x}\\\\##

    ##5^{log x}+5^{log x}=50\\\\##

    ##10^{log_{10} x}=50\Rightarrow x=50##


    They are common logarithms with base 10. I think you misunderstood it as base=e.
     
    Last edited: May 29, 2014
  11. May 29, 2014 #10

    ehild

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    Correct so far.
    That is wrong.

    ##2(5^{log x})≠10^{log x}##.

    From [itex]10^{\log x \log 5}+10^{\log 5 \log x}=50[/itex], [itex]10^{\log x \log 5}=25=5^2[/itex]. Take the logarithm of both sides.

    Some people use "log" for natural logarithm, others use it for base-10 logarithm. You need to clarify.

    ehild
     
  12. May 29, 2014 #11
    [itex]

    \log x \log 5=2 \log 5\\

    \log x=2 \Rightarrow x=100\\
    [/itex]
    Am I right?
     
  13. May 29, 2014 #12

    ehild

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    Yes!!!!! But check it. Plug-in x=100 into the original equation.

    ehild
     
  14. May 29, 2014 #13

    tms

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    Not quite. If
    $$ \log x\ \log 5 = 2\ \log 5\ \log x$$
    then if you divide both sides by ##\log x\ \log 5## you get
    $$ 1 = 2. $$
    And if you plug ##x = 100## into the original equation, you get something much larger than 50, or even 100.

    The right-hand side of
    $$ \log x\ \log 5 = 2\ \log 5\ \log x$$
    is right, but you need something else on the left-hand side.
     
    Last edited: May 29, 2014
  15. May 29, 2014 #14

    ehild

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    Nobody said that equation was true. The original equation was

    ##x^{log 5}+5^{log x}=50## . log means base-10 logarithm. It came out that

    ##\log x \log 5 = 2\log 5##


    Substitute back x=100: log(5)≈0.699, 100log(5)=25, 5log100=25...


    ehild
     
  16. May 29, 2014 #15

    Borek

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    Actually as long as you are looking for a symbolic representation of x what is the logarithm base doesn't matter.
     
  17. May 29, 2014 #16

    ehild

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    Borek,

    the problem has been solved correctly (Post #11). And the numerical value of x was needed. tms is wrong.

    ehild
     
  18. May 29, 2014 #17

    D H

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    It certainly does matter. The solution to ##x^{\log_b 5} + 5^{\log_b x} = 50## is x=4 if the base is 2, x=100 if the base is 10, and x=10000 if the base is 100.

    It is true that ##5^{\log_b x} = x^{\log_b 5}##, regardless of the base.
     
  19. May 29, 2014 #18
    Actually numerical value was needed. My teacher gave me four options a)25 b)50 c)75 d)100.


    He told me that simply ##\log x## means ##\log_{10} x##.

    And ##\ln x ## is used for ##\log_e x##.
     
  20. May 29, 2014 #19

    tms

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    It certainly seemed that you said it was correct when the OP asked if it was correct and you said yes. In retrospect you were saying the numerical answer was correct. I took the unadorned log to be a natural log and so got a very different answer.
     
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