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Thread: Need help with exponents

  1. #1
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    Need help with exponents

    I dont remember how to do this problem and being summer, I dont have my textbook to help.

    Solve: 4^(x+2)=9^(x-1)
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  2. #2
    Moo
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    Hello !
    Quote Originally Posted by ceasar_19134 View Post
    I dont remember how to do this problem and being summer, I dont have my textbook to help.

    Solve: 4^(x+2)=9^(x-1)
    Just use logarithms.

    Remember this property : $\displaystyle \ln(a^b)=b \ln(a)$

    $\displaystyle \ln(4^{x+2})=\ln(9^{x-1})$

    $\displaystyle (x+2) \ln(4)=(x-1) \ln(9)$

    Can you do the rest ? (you can simplify again in the end by writing $\displaystyle 4=2^2 \quad \& \quad 9=3^2$)
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  3. #3
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    I still cant figure it out, sorry. All of the other problems I have to do have the same base or can be simplified to have the same base.
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  4. #4
    Moo
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    Quote Originally Posted by ceasar_19134 View Post
    I still cant figure it out, sorry. All of the other problems I have to do have the same base or can be simplified to have the same base.
    Whatever base you choose, the rules are the same

    $\displaystyle \ln$ denotes the natural logarithm and is like an *universal* base :
    $\displaystyle \ln(e)=1$

    Continuing from $\displaystyle (x+2) \ln(4)=(x-1) \ln(9)$ :

    $\displaystyle x \ln(4)+2 \ln(4)=x \ln(9)-\ln(9)$

    Group the x together :

    $\displaystyle x \ln(9)-x \ln(4)=2 \ln(4)+\ln(9)$

    $\displaystyle x(\ln(9)-x \ln(4))=2 \ln(4)+\ln(9)$

    $\displaystyle \implies x=\frac{2 \ln(4)+\ln(9)}{\ln(9)-\ln(4)} \quad (1)$

    Now, if you take the logarithm in base 2, you'll have $\displaystyle \log_2(4)=2$ because $\displaystyle 4=2^2$. Actually, it's $\displaystyle \log_2(4)=\log_2(2^2)=2 \underbrace{\log_2(2)}_{=1}=2$.

    But in a common logarithm, you have :
    $\displaystyle \ln(4)=\ln(2^2)=2 \ln(2)$, same goes for $\displaystyle \ln(9)=\ln(3^2)=2 \ln(3)$

    Substituting in (1) :
    $\displaystyle x=\frac{4 \ln(2)+2 \ln(3)}{2 \ln(3)-2 \ln(2)}$

    Simplify by 2 :

    $\displaystyle \boxed{x=\frac{2 \ln(2)+\ln(3)}{\ln(3)-\ln(2)}}$


    You gotta learn how to use the natural logarithm. Using bases just sometimes avoid difficulties things easier, but in exercises like this one, it brings difficulties !
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  5. #5
    Super Member Showcase_22's Avatar
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    I did it a different way:

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