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Thread: Logs???

  1. #1
    daniella
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    Exclamation Logs???

    how do i use the "change of base" formula to solve

    1 a) 3^2x=8
    it is the exponent 2x
    2 d) 5^x-2=150
    it is to the exponent x-2

    Thank You!
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  2. #2
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    Quote Originally Posted by daniella View Post
    how do i use the "change of base" formula to solve

    1 a) 3^2x=8
    it is the exponent 2x
    $\displaystyle 3^{2x}=8=2^3$
    Thus,
    $\displaystyle 2x=3$
    $\displaystyle x=1.5$
    2 d) 5^x-2=150
    it is to the exponent x-2
    $\displaystyle 5^{x-2}=150$
    Thus,
    $\displaystyle \log_5 150=x-2$
    $\displaystyle x=2+\log_5 150$
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  3. #3
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    Hello, daniella!

    You're expected to know that $\displaystyle b^x\:=\:a$ is equivalent to $\displaystyle x \:=\:\log_ba$
    . . and the Base-change formula: .$\displaystyle \log_aX \;=\;\frac{\log_bX}{\log_ba}$


    How do i use the "change of base" formula to solve:

    $\displaystyle 1\:a)\;3^{2x}\:=\:8$

    $\displaystyle 2\:d)\;5^{x-2}\:=\:150$

    "Change of Base" formula: .$\displaystyle \log_aX \:=\:\frac{\log_bX}{\log_ba}$


    1a) We have: .$\displaystyle 3^{2x} \:=\:8$ . . . which is equivalent to: .$\displaystyle 2x \:=\:\log_38$

    Change-of-base: .$\displaystyle 2x\:=\:\frac{\log8}{\log3} $

    Therefore: .$\displaystyle x \;=\;\frac{\log8}{2\cdot\log3} \;\approx\;0.9464$


    2d) We have: .$\displaystyle 5^{x-2}\;=\;150$ . . . which is equivalent to: .$\displaystyle x-2 \;=\;\log_5150$

    Change-of-base: .$\displaystyle x-2\;=\;\frac{\log150}{\log5}$

    Therefore: .$\displaystyle x \;=\;\frac{\log150}{\log5} + 2 \;\approx\;5.1133$

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