anonymous
  • anonymous
1. Create a graph of the pH function either by hand or using technology. Locate your graph where the pH value is 0 and where it is 1. You may need to zoom in on your graph.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
Any ideas? IrishBoy :)
anonymous
  • anonymous
Before any plotting, let's create a table, |dw:1438978318051:dw|

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anonymous
  • anonymous
Ok :) I see. Now what do we do?
anonymous
  • anonymous
take any numbers of values, then plot them on a graph |dw:1438978459965:dw|
anonymous
  • anonymous
Oh! Cant we use graphing technology? Like fooplot.com?
anonymous
  • anonymous
That is something you would know, is this a question for school or what?If so you should ask your teacher if you can use a graph online
anonymous
  • anonymous
You can:)
anonymous
  • anonymous
Ok the question says either by hand or by technology, so you're free to use any online graphing softwares
anonymous
  • anonymous
I was just wondering, if you could just use graphing technology instead of drawing it to make it more clear
anonymous
  • anonymous
So what would the graph look like?
anonymous
  • anonymous
anonymous
  • anonymous
|dw:1438979039696:dw| It would look something like this, search "graph -log(x)" on google and it will generate a graph for you
anonymous
  • anonymous
Alright thanks. Now what about this 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.
mathmate
  • mathmate
First we will use the identity \(\Large y=log_b x~~means~~ x=b^y\) so if \(p(t)=y=-log10(t)=log10(1/t)\) then \(1/t=10^y\) or \(t=10^{1/y}=10^{-y}\) If\( y=0.5~~then~~ t=10^{-0.5}\)
anonymous
  • anonymous
|dw:1438979139010:dw| Isn't it -log10(2)=-2??
anonymous
  • anonymous
Am I wrong?
anonymous
  • anonymous
\[p(t)=-\log(t)\] If you shift your t by 0.5, your graph will also change, I don't see how you could use the same graph, you would need to shift it slightly\[p(t+0.5)=-\log(t+0.5)\]
anonymous
  • anonymous
yes you are wrong, what I mean by in that table is \[-\log_{10}(2)=-0.3010\]
anonymous
  • anonymous
Last one! The pool company developed new chemicals that transform the pH scale. Using the pH function p(t) = −log10t as the parent function, explain which transformation results in a y-intercept and why. You may graph by hand or using technology. Use complete sentences and show all translations on your graph. p(t) + 1 p(t + 1) −1 • p(t)
mathmate
  • mathmate
Check: p(10^-0.5)=-log10(10^-0.5)=-(-0.5)=0.5
mathmate
  • mathmate
g(x)=f(x)+k will translate the graph of f(x) vertically (shifts up by k) g(x)=f(x-h) will translate a graph of f(x) horizontally to the right by "h". which implies g(x)=f(x+h) will translate graph of f(x) to the left by h. g(x)=-f(x) will flip the graph about the x-axis. So make your choice according to the requirements.
anonymous
  • anonymous
um the first one?
anonymous
  • anonymous
I dont understand how to graph #1?
mathmate
  • mathmate
You have already graphed p(t) at the very beginning.|dw:1438979941325:dw| You will have to figure out which translation will result in a y-intercept!
anonymous
  • anonymous
ok so -log10(0)=0 -log10(1)=-1
anonymous
  • anonymous
I understand:) Whoops! Ok let's got back to #3 bcuz i dont understand that one
anonymous
  • anonymous
The pool company developed new chemicals that transform the pH scale. Using the pH function p(t) = −log10t as the parent function, explain which transformation results in a y-intercept and why. You may graph by hand or using technology. Use complete sentences and show all translations on your graph. p(t) + 1 p(t + 1) −1 • p(t)
mathmate
  • mathmate
no, log(1)=0 (to any base) and log(0) does not exist.
anonymous
  • anonymous
oh! Sorry !!
mathmate
  • mathmate
Sorry, gtg, I'll continue later if no one gave you help in the mean time!
anonymous
  • anonymous
Thanks so much:)

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