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Homework Help: Finding the value of an exponential given that e^a=10 and e^b=4

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that e^a=10 and e^b=4. Find the value of


    2. Relevant equations


    3. The attempt at a solution

    So far I've only gotten to the above equation rearranged to:

    [itex]ln(b) loga(e)=loga(b)/[/itex]. I'm unsure as to where I'm supposed to go from here...
  2. jcsd
  3. Oct 8, 2012 #2
    A useful relation is loga b = logeb / logea.
    How is logeb related to eb?
  4. Oct 8, 2012 #3
    Hmm, you can get e^b from it by going exp(exp(ln(b))), but I'm kinda stuck. I get up to this stage:



    I can re-arrange so that I get exp(loga(b))=(exp(lnb))1/lna)=b^1/lna, then I'm unsure where to go as I dont know what to do with the 1/lna.
  5. Oct 8, 2012 #4
    EDIT: That doesn't seem useful for the problem here. What you should use is b = ln eb.

    Find ln a just the way you found ln b and substitute it here.

    You don't need to do this. :smile:
  6. Oct 8, 2012 #5
    Wait, I just realized that all I had to do was solve for a and b in the original equations and plug it in. Doh! Thanks for the help though :)
  7. Oct 8, 2012 #6


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    And if you want to write it in terms of 4 and 10 then it's [itex] \frac{\ln \ln 4}{ \ln \ln 10}[/itex]
  8. Oct 8, 2012 #7


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    (Do NOT put the HTML "sub" tags inside "tex" tags, either do not use "tex" or use underscores, "_", in "tex".)
    For any x, eln(x)= x so that [itex]a^x= e^{ln(a^x)}= e^{x ln(a)}[/itex]. Similarly, if [itex]y= log_a(b)[/itex], then [itex]b= a^y[/itex] so that [itex]b= e^{ln a^y}= e^{y ln a}[/itex]. Then [itex]ln(b)= y ln(a)[/itex] so that [itex]y= ln_a(b)= ln(b)/ln(a)[/itex].
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