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Thread: Quick Algebra question (logs and stuff)

  1. #16
    Junior Member reynardin's Avatar
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    i get a negative number every time i try to solve...
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  2. #17
    Eater of Worlds
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    Let's do it this way then.

    Rewrite $\displaystyle e^{\frac{ln(3/5)t}{5}}=(3/5)^{\frac{t}{5}}$

    $\displaystyle 50=100(3/5)^{\frac{t}{5}}$

    $\displaystyle 1/2=(3/5)^{\frac{t}{5}}$

    $\displaystyle ln(1/2)=\frac{t}{5}ln(3/5)$

    $\displaystyle t=\frac{5ln(1/2)}{ln(3/5)}=6.78$
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  3. #18
    Junior Member reynardin's Avatar
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    ooooooh
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  4. #19
    Eater of Worlds
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    Quote Originally Posted by reynardin View Post
    wait, so is K supposed to equal the $\displaystyle

    \frac{\ln(\frac{3}{5})}{5}=\frac{\ln\bigg(\bigg(\f rac{5}{3}\bigg)^{-1}\bigg)}{5}=\frac{-1\ln(\frac{5}{3})}{5}
    $?
    Yes. That is the decay constant you need.
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  5. #20
    Junior Member reynardin's Avatar
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    ok, i suppose I get it now.

    Next problem.

    $\displaystyle \log_3(x)-\log_4(7)=\log_7(5)$

    I put the equation into logs of base 10, but I don't know what to do after that.
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  6. #21
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    Quote Originally Posted by reynardin View Post
    ok, i suppose I get it now.

    Next problem.

    $\displaystyle \log_3(x)-\log_4(7)=\log_7(5)$

    I put the equation into logs of base 10, but I don't know what to do after that.
    Of base 10 ? o.O

    Put it in the exponential logarithm... $\displaystyle \log_x (y)=\frac{\ln(y)}{\ln(x)}$


    ---> $\displaystyle \frac{\ln x}{\ln 3}-\frac{\ln 7}{\ln 4}=\frac{\ln 5}{\ln 7}$

    --> $\displaystyle \ln x-\frac{\ln 7 \cdot \ln 3}{\ln 4}=\frac{\ln 5 \cdot \ln 3}{\ln 7}$

    $\displaystyle \ln x=\frac{\ln 5 \cdot \ln 3}{\ln 7}+\frac{\ln 7 \cdot \ln 3}{\ln 4}$

    ---> $\displaystyle x=exp\left(\frac{\ln 5 \cdot \ln 3}{\ln 7}+\frac{\ln 7 \cdot \ln 3}{\ln 4}\right) \approx 11.59699220980753$
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  7. #22
    Junior Member reynardin's Avatar
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    EH, thats what i mean >.<
    lol
    as in, log base ten of x
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  8. #23
    Junior Member reynardin's Avatar
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    BLAH ignore last post
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  9. #24
    Junior Member reynardin's Avatar
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    what does the exp mean?
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  10. #25
    Moo
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    Quote Originally Posted by reynardin View Post
    what does the exp mean?
    $\displaystyle exp(\dots)$ is like $\displaystyle e^{\dots}$

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  11. #26
    Junior Member reynardin's Avatar
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    ah ok
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  12. #27
    Junior Member reynardin's Avatar
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    anyways,
    I LOVE YOU GUYS!

    done for now =D
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  13. #28
    Junior Member reynardin's Avatar
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    Quote Originally Posted by galactus View Post
    Let's do it this way then.

    Rewrite $\displaystyle e^{\frac{ln(3/5)t}{5}}=(3/5)^{\frac{t}{5}}$

    $\displaystyle 50=100(3/5)^{\frac{t}{5}}$

    $\displaystyle 1/2=(3/5)^{\frac{t}{5}}$

    $\displaystyle ln(1/2)=\frac{t}{5}ln(3/5)$

    $\displaystyle t=\frac{5ln(1/2)}{ln(3/5)}=6.78$
    btw, how does he get $\displaystyle e^{\frac{ln(3/5)t}{5}}=(3/5)^{\frac{t}{5}}$? I get that ln and e basically "cancel" each other out, but why is the 3/5 to the power of t/5?
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  14. #29
    Junior Member reynardin's Avatar
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    Disregard my last post, i figured it out.
    Quote Originally Posted by Moo View Post
    Hello,
    ...
    Put it in the exponential logarithm... $\displaystyle \log_x (y)=\frac{\ln(y)}{\ln(x)}$


    ---> $\displaystyle \frac{\ln x}{\ln 3}-\frac{\ln 7}{\ln 4}=\frac{\ln 5}{\ln 7}$

    --> $\displaystyle \ln x-\frac{\ln 7 \cdot \ln 3}{\ln 4}=\frac{\ln 5 \cdot \ln 3}{\ln 7}$

    $\displaystyle \ln x=\frac{\ln 5 \cdot \ln 3}{\ln 7}+\frac{\ln 7 \cdot \ln 3}{\ln 4}$

    ---> $\displaystyle x=exp\left(\frac{\ln 5 \cdot \ln 3}{\ln 7}+\frac{\ln 7 \cdot \ln 3}{\ln 4}\right) \approx 11.59699220980753$
    does multiplying an equation such as $\displaystyle \ln(x)=....$ by e make it into $\displaystyle x=e*...$?
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  15. #30
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by reynardin View Post
    Disregard my last post, i figured it out.

    does multiplying an equation such as $\displaystyle \ln(x)=....$ by e make it into $\displaystyle x=e*...$?
    For the last one

    $\displaystyle a^{bc}=(a^b)^c$

    So $\displaystyle e^{\ln(\frac{3}{5})\frac{t}{5}}=(e^{\ln(\frac{3}{5 })})^{\frac{t}{5}}=\bigg(\frac{3}{5}\bigg)^{\frac{ t}{5}}$

    and no multiplying by it does nothign

    raising e to each side power does

    for example $\displaystyle \ln(45x-220)=\ln(5+17x)$

    If you multiplied both sides by e would do nothing

    but raising e to eachside would give

    $\displaystyle e^{\ln*45x-220)}=e^{\ln(5+17x)}\Rightarrow{45x-220=5+17}$
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