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Solving Exponential Equations

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve [itex]2^{k-2}=3^{k+1}[/itex]

    2. Relevant equations

    3. The attempt at a solution
    [tex]\frac{2^{k}}{3^{k}} = 4 \cdot 3[/tex]
    [tex]\frac{2^{k}}{3^{k}} = 12[/tex]

    What do I do next to solve for K?
  2. jcsd
  3. Jan 26, 2012 #2


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    Homework Helper

    Use the fact that [tex]\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n[/tex] and also, what do you know about logarithms?
  4. Jan 27, 2012 #3


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    Science Advisor

    You could have used logarithms right from the start- [itex]log(2^{k-2})= log(3^{k+1})[/itex].

    In general, to solve an equation of the form f(x)= constant or f(p(x))= f(q(x)) you will need to use the inverse function to f. And the inverse of the exponential is the logarithm.
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