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Math Help - Logarithms, Please help

  1. #1
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    Logarithms, Please help

    Couple of questions dealing with logarithms

    I just need to see how they are done, so if you could work them step by step.


    If logb3=.234 and logb2=.105, find logb6 .


    Solve for x: log(x-5) + log(x-10)=3


    438=200e.25x


    A=P(1+(r/n))nt for t



    4e3x=20

    Thanks in advance!!
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  2. #2
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    This isn't really a calculus problem, but seeing that it's your first post, I'll help you anyway.

    For the first problem, we may use the fact that \log xy = \log x +\log y. Here,

    \log_b 6 =\log_b 2\cdot 3=\log_b 2+\log_b 3.

    For the second problem, we may deduce a further equality by raising 10 to the power of both sides:

    \begin{aligned}<br />
10^{\log(x-5)+\log(x-10)}&=10^3\\<br />
10^{\log(x-5)}\cdot 10^{\log(x-10)}&=1000\\<br />
(x-5)(x-10)&=1000.<br />
\end{aligned}

    Two of the remaining problems can be solved by taking the logarithm of both sides.
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  3. #3
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    Thanks for the help on the other ones!!

    Quote Originally Posted by Scott H View Post
    Two of the remaining problems can be solved by taking the logarithm of both sides.


    438=200e^{0.25x}



    A=P(1+(r/n))^{nt} for t


    I know I need to use logarithms for these but every time I work them I get a different answer. I just need to see them worked once, then I usually get 'em.
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  4. #4
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    438=200e^{0.25x} \implies 0.25x = \ln{\frac{438}{200}} \implies x = 4\ln{\frac{438}{200}}



    A=P(1+(r/n))^{nt} \implies \frac{A}{P}=(1+(r/n))^{nt} \implies \log_{1+(r/n)}\frac{A}{P}=nt \implies t = \frac{\ln{\frac{A}{P}}}{n \cdot \ln({1+(r/n)}) }
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