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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
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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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0In 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
 one year ago
Best ResponseYou've already chosen the best response.0or, alternatively, you can say, A = log\(\large \rm _{_B}\)(C) A = log\(\large \rm _{_B}\)(C) C = B\(\Large ^{\rm _{^{{\bf \LARGE}A}}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This 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 \[y = \log_{b} x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@fashionismybesty are you still there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes 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
 one year ago
Best ResponseYou've already chosen the best response.0Where 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
 one year ago
Best ResponseYou've already chosen the best response.0And NO log (2) is not 2. Log 2 would be 0.3010299957....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So anyway let us solve this problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x = b^y\] Is a exponential function. We have \[y = \log (t)\] Remember that \[\log_{x} (y)^z = z \log_{x} (y)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, which means that log(t^1)=y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So in this case then \[y=−\log(t)\] would become \[y = \log (t)^1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then using \[b^y = x\] and \[y = \log_{b} x\] We get, \[10^y = t^1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now remember, \[t^{1} = \frac{ 1 }{ t }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But when you graph the exponential function, isn't it supposed to be a reflection of the logarithm over the yaxis? It isn't when I graph \[10^y=t^1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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
 one year ago
Best ResponseYou've already chosen the best response.0Sorry about taking a long time, my computer is glitching when i use equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It'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
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.0That 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, that makes a lot of sense. Thanks again!
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