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Homework Help: Exponents with different bases

  1. Aug 26, 2008 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    basically all I want to know is what is the best way to go about solving a problem like this.

    3. The attempt at a solution
    I know how to convert a problem as such 4^(3x)=10 using. [log(<base>4,10)]/3 = X. however I need some help with addition and subtraction in exponents with one base equaling another number with a different base.
  2. jcsd
  3. Aug 26, 2008 #2
    I don't see what the problem is. I mean on the LHS you have factors of 2 and on the RHS you have factors of 3, so you really can't simplify further in a sense.

    The best way to go is using logarithms. It should be clear you won't find a "nice" value for x (obviously no integer solutions).
  4. Aug 26, 2008 #3
    Well, as for the addition and subtraction, it seems you have the right idea by dividing that logarithm of base 4 by 3. It's the same idea, algebra is algebra. You can look at the log as a variable to help get your mind around it. (x+3 instead of log(<base>3,6)+3) And as for the different bases, there is a handy formula called the Change-of-Base formula! It will be your best friend with logarithms. Here is a quick copy and paste of it, you can find more if you search it using yahoo or google.

    [tex]log_a(x)[/tex] can be converted to base 'b' by the formula

    (sorry, not too familiar with LaTeX, but I hope you get the idea. As I said, you can just search for the formula. Best of luck!)
  5. Aug 27, 2008 #4


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    Science Advisor

    It doesn't matter what base you use: if 4x+2= 9 then log(4x+2)= (x+2)log 4= log 9 where "log" can be to any base. I presume your calculator has both base 10 and natural logarithms so use those and solve for x.
  6. Aug 27, 2008 #5
    Using logs I've found that X=(log9/log4)-2 = log9-log4-2. Is this correct?
  7. Aug 27, 2008 #6


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    Homework Helper

    No! your final statement

    \frac{\log 9}{\log 4} - 2 = \log 9 - \log 4 - 2

    is not correct.

    It is true that

    \log \left( \frac 9 4 \right) = \log 9 - \log 4

    but that is not what your answer involves.
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