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Math Help - natural logs proof eqn

  1. #1
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    natural logs proof eqn

    can somebody help.I have no idea where to start with this equation. I presume its actually quite easy but some concept I have forgotton. Q
    If x satisfies 3^x.4^{2x+1}=6^{x+2}, show that
    x=\frac{ln9}{ln8},,,,
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by oxrigby View Post
    can somebody help.I have no idea where to start with this equation. I presume its actually quite easy but some concept I have forgotton. Q
    If x satisfies 3^x.4^{2x+1}=6^{x+2}, show that
    x=\frac{ln9}{ln8},,,,
    first, take the natural logarithm of both sides:

    \ln\left(3^x4^{2x+1}\right)=\ln\left(6^{x+2}\right  )

    Now recall that \ln(ab)=\ln(a)+\ln(b)

    Thus, \ln\left(3^x4^{2x+1}\right)=\ln\left(6^{x+2}\right  )\implies \ln\left(3^x\right)+\ln\left(4^{2x+1}\right)=\ln\l  eft(6^{x+2}\right)

    Now recall that \ln(a^b)=b\ln a

    So, \ln\left(3^x\right)+\ln\left(4^{2x+1}\right)=\ln\l  eft(6^{x+2}\right)\implies x\ln\left(3\right)+(2x+1)\ln\left(4\right)=(x+2)\l  n\left(6\right)

    Do you think you can take it from here and try to solve for x?

    --Chris
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  3. #3
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    Quote Originally Posted by oxrigby View Post
    can somebody help.I have no idea where to start with this equation. I presume its actually quite easy but some concept I have forgotton. Q
    If x satisfies 3^x.4^{2x+1}=6^{x+2}, show that
    x=\frac{ln9}{ln8},,,,
    Take logs,
    \ln (3^x \cdot 4^{2x+1} ) = \ln 6^{x+2}
    Thus,
    \ln 3^x + \ln 4^{2x+1} = \ln 6^{x+2}
    Therefore,
    x\ln 3 + (2x+1) \ln 4 = (x+2)\ln 6
    We get,
    x\ln 3 + 2x\ln 4 + \ln 4 = x\ln 6 + 2\ln 6
    Thus,
    x(\ln 3 + 2\ln 4 - \ln 6) = 2\ln 6 - \ln 4
    This gives,
    x\cdot \ln \frac{3\cdot 2^4}{6} = \ln \frac{6^2}{4}
    Thus we get,
    x\ln 8 = \ln 9 \implies x = \frac{\ln 9}{\ln 8}
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