Pre-Cal :)))

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Create a graph of the pH function either by hand or using technology. Locate on your graph where the pH value is 0 and where it is 1. You may need to zoom in on your graph.
  • phi
does p(t) = −log10t. mean \[ p(t)= - \log_{10} t \] or \[ p(t) = −\log(10t)\]?
The 1st one lol!

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Other answers:

  • phi
ok, I just googled it. Yes the first way. \[ p(t)= - \log_{10} t \] pick some values for t. I would pick powers of 10 (because it is easy to "take the log" ) thus let t= 0.001 (i.e. t= 10^-3 t= 0.01, or t= 10^-2 t= 0.1 or t= 10^-1 t= 1, t= 10^0 t=10 , t= 10^1
Ok, then how would u graph it?
  • phi
first, what is the log of 10^-3 ? \[ \log_{10} \left(10^{-3} \right) = \]
0.001?
  • phi
10^-3 is 0.001 the log of 0.001 is the exponent (assuming we use base 10) in other words \[ \log_{10}(0.001) = \log_{10}\left( 10^{-3}\right) \] what is the exponent ?
-3
  • phi
ok. \[ p(t)= - \log_{10} t \\= -1 \cdot \log(0.001) \\= -1 \cdot -3 = 3 \]
  • phi
so (0.001, 3) is a point on the graph hopefully you can figure out the following points (0.01,2) (0.1, 1) (1,0) (10,-1) (100,-2) (1000,-3)
yes, I think I can. How could I graph this using technology?
  • phi
if you have geogebra , you type in the equation
I dont have that/ know what that is lol :)
I have desmos.com but I wouldnt know how to type that in
Anyways, I'll figure it out:) No worries!! This is the 2nd part. 2. The pool maintenance man forgot to bring his logarithmic charts, and he needs to raise the amount of hydronium ions, t, in the pool by 0.50. To do this, he can use the graph you created. Use your graph to find the pH level if the amount of hydronium ions is raised to 0.50. Then, convert the logarithmic function into an exponential function using y for the pH.
  • phi
just type in y= -log(x)
oH! Hahaha thanks :)
What about the 2nd part of the question?
  • phi
zoom in on your graph, and click on the curve when it has y=0.5 what do you get for the x value ?
I got (0.316,0.5)
  • phi
for ** convert the logarithmic function into an exponential function using y for the pH.*** y= - log(t) or, after multiplying by -1 on both sides: -y = log(t) now make each side the exponent of 10 \[ 10^{-y} = 10^{\log t} \] the right side becomes t \[ 10^{-y} = t \\ t= 10^{-y} \]
  • phi
yes, 0.316 is good using the exponential equation \[ t= 10^{-0.5} = \frac{1}{\sqrt{10}} = 0.316228\]
So, we just converted it right? And that would be the answer to the question?
  • phi
yes, the work to find the exponential form is up above. and I used it to check your answer... and as you see we get the same number (rounded of course)
Yes! Thanks so much<3

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