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Need help with Logs

  1. Apr 22, 2016 #1
    1. The problem statement, all variables and given/known data
    (1/25)^x+3=125^x+3

    2. Relevant equations
    I have used log on both sides but keep getting an incorrect answer.

    3. The attempt at a solution
    How would solve this type of equation. I set the bases both to 5, make them equal to each other and i cant get the right answer.
     
    Last edited by a moderator: Apr 30, 2016
  2. jcsd
  3. Apr 22, 2016 #2

    Math_QED

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    Is the + 3 in the exponent? If not you can substract it from both sides.
     
    Last edited by a moderator: Apr 30, 2016
  4. Apr 22, 2016 #3

    SammyS

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    If the exponent DOES include the + 3 , then the whole exponent should be enclosed in parentheses.

    We follow standard Order of Operations here at PF .
     
  5. Apr 22, 2016 #4
    yes the +3 is in the exponent. So i should put the whole thing in parentheses.
     
  6. Apr 22, 2016 #5

    Math_QED

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    Make the substitution k = x + 3, then take the log of both sides.
     
  7. Apr 22, 2016 #6
  8. Apr 22, 2016 #7
    what is k. Is that the (1/25)?
     
  9. Apr 22, 2016 #8

    Math_QED

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    So the equation is:

    (1/25)^(x+3) = 125^(2x) ?

    EDIT: The equation you initially posted seems to be not the same as the one you wrote down here. Never mind what I said about substitution.
     
  10. Apr 22, 2016 #9
    yes that is the equation my apologies.
     
  11. Apr 22, 2016 #10

    Mark44

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    What you wrote in the first post is different from the image in your later post.
    The equation apparently is ##(\frac 1 {25})^{x + 3} = 125^{2x}##

    Link to our LaTeX primer: https://www.physicsforums.com/help/latexhelp/
     
  12. Apr 22, 2016 #11
    the equation (1/25)^(x+3)= 125^2X thats the equation on the paper.
     
  13. Apr 22, 2016 #12
    sorry i didnt click on the link before i replied i will use the Latex thing next time.
     
  14. Apr 22, 2016 #13

    Math_QED

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    Regarding your equation. You must have seen that 1/25 and 125 can be written as 5^(-2) and [itex]5^3[/itex]
    Rewrite your equation with this. Post what you get then.
     
  15. Apr 22, 2016 #14

    SammyS

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    If you write expression in that "in-line" sort of format, please use adequate parentheses.

    For example:
    (1/25)^(x+3)= 125^(2x)​

    .
     
  16. Apr 22, 2016 #15
    I make it 5^(-2)(X+3)=5^(3)(2x) and this is where I think I mess up. I multiply the numbers in parentheses
     
  17. Apr 22, 2016 #16

    Math_QED

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    Now use: [itex](a^b)^c [/itex] = a^(bc)
     
  18. Apr 22, 2016 #17

    SammyS

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    f(x) = 5x is a one-to-one function.

    Therefore, just equate the arguments of the two sides.
     
  19. Apr 29, 2016 #18
    Sorry for starting up this thread again, but what do I put in that equation. Like is 5=A, I don't understand what numbers to plug in.
     
  20. Apr 29, 2016 #19

    Math_QED

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    You can ignore that. Once you have the expression in #16, you can say make an equation where you say the exponents are equal. (Do in fact you take the 5 bases log of what's left and right from the equality sign. Once you do that, you get an easy equation in x to solve. Let us know when you found the answer :)
     
  21. Apr 30, 2016 #20

    NascentOxygen

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    Yes, but incorporating the extra necessary parentheses as SammyS corrected for you in post #18.

    Please apply more care in your future threads. Careless mistakes and omissions in presentation don't give confidence that you are valuing the assistance of forum helpers.
     
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