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Math Help - 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 : \ln(a^b)=b \ln(a)

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

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

    Can you do the rest ? (you can simplify again in the end by writing 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

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

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

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

    Group the x together :

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

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

    \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 \log_2(4)=2 because 4=2^2. Actually, it's \log_2(4)=\log_2(2^2)=2 \underbrace{\log_2(2)}_{=1}=2.

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

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

    Simplify by 2 :

    \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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