anonymous
  • anonymous
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
Algebra
chestercat
  • chestercat
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SolomonZelman
  • SolomonZelman
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\(\Large ^{\rm _{^A}}\)
SolomonZelman
  • SolomonZelman
or, alternatively, you can say, A = -log\(\large \rm _{_B}\)(C) -A = log\(\large \rm _{_B}\)(C) C = B\(\Large ^{\rm _{^{{\bf \LARGE-}A}}}\)
anonymous
  • anonymous
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 \[y = \log_{b} x\]

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anonymous
  • anonymous
@fashionismybesty are you still there?
anonymous
  • anonymous
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
anonymous
  • anonymous
anonymous
  • anonymous
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.
anonymous
  • anonymous
And NO log (2) is not 2. Log 2 would be 0.3010299957....
anonymous
  • anonymous
So anyway let us solve this problem.
anonymous
  • anonymous
Ok
anonymous
  • anonymous
\[x = b^y\] Is a exponential function. We have \[y = -\log (t)\] Remember that \[\log_{x} (y)^z = z \log_{x} (y)\]
anonymous
  • anonymous
yes, which means that log(t^-1)=y
anonymous
  • anonymous
So in this case then \[y=−\log(t)\] would become \[y = \log (t)^-1\]
anonymous
  • anonymous
So then using \[b^y = x\] and \[y = \log_{b} x\] We get, \[10^y = t^-1\]
anonymous
  • anonymous
i mean \[t ^{-1}\]
anonymous
  • anonymous
now remember, \[t^{-1} = \frac{ 1 }{ t }\]
anonymous
  • anonymous
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 \[10^y=t^-1\]
anonymous
  • anonymous
t^-1=10^y I mean
anonymous
  • anonymous
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 \[t = 10^{-y}\]
anonymous
  • anonymous
Sorry about taking a long time, my computer is glitching when i use equation
anonymous
  • anonymous
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?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
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
anonymous
  • anonymous
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)
anonymous
  • anonymous
Oh, that makes a lot of sense. Thanks again!

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