1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    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


    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?