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anonymous

  • one year ago

Convert the logarithmic function into an exponential function using y for the pH. p(t) = −log10t. I really need help with this, any help is appreciated

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  1. SolomonZelman
    • one year ago
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    In general: A = -log\(\large \rm _{_B}\)(C) A = log\(\large \rm _{_B}\)(C\(^{-1}\)) A = log\(\large \rm _{_B}\)(1/C) 1/C = B\(\Large ^{\rm _{^A}}\)

  2. SolomonZelman
    • one year ago
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    or, alternatively, you can say, A = -log\(\large \rm _{_B}\)(C) -A = log\(\large \rm _{_B}\)(C) C = B\(\Large ^{\rm _{^{{\bf \LARGE-}A}}}\)

  3. anonymous
    • one year ago
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    This is simple. I also did this for FLVS. So now you will have the equation. y= −log (t) because \[\log_{10} \] is the same as log. Now remember this? \[b^y = x\] and \[y = \log_{b} x\]

  4. anonymous
    • one year ago
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    @fashionismybesty are you still there?

  5. anonymous
    • one year ago
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    Yes I do remember this (I'm also on flvs) and I'm just confused because logarithms usually equal x and now it equals y

  6. anonymous
    • one year ago
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    @whatdoesthismean

  7. anonymous
    • one year ago
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    Where did you hear logarithms equal x? it could equal a,b,c,d,e,f,d,h, and etc. Logarithms like log (1) would equal 1, and so on.

  8. anonymous
    • one year ago
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    And NO log (2) is not 2. Log 2 would be 0.3010299957....

  9. anonymous
    • one year ago
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    So anyway let us solve this problem.

  10. anonymous
    • one year ago
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    Ok

  11. anonymous
    • one year ago
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    \[x = b^y\] Is a exponential function. We have \[y = -\log (t)\] Remember that \[\log_{x} (y)^z = z \log_{x} (y)\]

  12. anonymous
    • one year ago
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    yes, which means that log(t^-1)=y

  13. anonymous
    • one year ago
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    So in this case then \[y=−\log(t)\] would become \[y = \log (t)^-1\]

  14. anonymous
    • one year ago
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    So then using \[b^y = x\] and \[y = \log_{b} x\] We get, \[10^y = t^-1\]

  15. anonymous
    • one year ago
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    i mean \[t ^{-1}\]

  16. anonymous
    • one year ago
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    now remember, \[t^{-1} = \frac{ 1 }{ t }\]

  17. anonymous
    • one year ago
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    But when you graph the exponential function, isn't it supposed to be a reflection of the logarithm over the y-axis? It isn't when I graph \[10^y=t^-1\]

  18. anonymous
    • one year ago
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    t^-1=10^y I mean

  19. anonymous
    • one year ago
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    But we're not done. \[t^{-1} = 10^y\] and \[\frac{ 1 }{ t } = 10^y\] just don't work. But wait, we have a way. remember, \[t^{-1} = 10^y\] is equivalent to \[t = 10^{-y}\]

  20. anonymous
    • one year ago
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    Sorry about taking a long time, my computer is glitching when i use equation

  21. anonymous
    • one year ago
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    It's fine and thanks for the help so much! for the y to become negative, you just multiply it by the -1 exponent right?

  22. anonymous
    • one year ago
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    yeah

  23. anonymous
    • one year ago
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    I'm still a little bit confused because the graph of the exponential functions is supposed to be the reflection over the y axis of the logarithm and it isn't

  24. anonymous
    • one year ago
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    That is because this is the exponential function FORM of the logarithmic function p(t). Exponential functions are the inverse of the logarithmic function, so the exponential function of p(t) would be y = 10^x. THAT would be the reflection over the y - axis of p(t). In this question we need to just find the FORM of p(t), not the exponential function (or inverse)

  25. anonymous
    • one year ago
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    Oh, that makes a lot of sense. Thanks again!

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