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college algebra homework help please

So the chapter is on logarithmic functions and I'm having trouble with this problem:

An ulcer patient has been told to avoid acidic foods. If he drinks coffee, with a pH of 5.0, it bothers him, but he can tolerate both tap water, with a pH of 5.8, and milk, with a pH of 6.9.
a) What pH will the half coffee-half milk mixture have? ---> I already got this answer, which was a pH of 5.295

b) In order to make 18 oz of a milk-coffee drink with a pH of 5.8, how many ounces of each are required?

I need help with part b^

Any and all help is appreciated!

November 2, 2014

6 Comments • Newest first

FrozenCubez

The equation should look like :
[H+] = (1-a)(x) + (a)(y)
Where:
5.8 = -log [H+]
5.0 = -log x
6.9 = -log y
(1-a) = proportion of x
a = proportion of y

And when you solve for a, figure out how much ounces that is.

Should be enough info for you to solve it

Reply November 4, 2014 - edited
Avatar

[quote=Valued]@Avatar: can you pm me it? or will it block it too? if you dont mind can you write it out by words? idk lol[/quote]

well for equation in part a you had something like y equals half times 10 to the power of negative ph of coffee + half times 10 to the power of negative ph of milk. So if you sub the ph's you should get the answer for a. Now just replace the first half with x and the other half with 1-x and sub in y. You can expand the (1-x) part, collect the terms with x and solve for x.

Edit: my bad it should be -log(y) on the left side of the equation. so take the negative of the logarithm of the ph they gave you in b

Reply November 3, 2014 - edited
Valued

@Avatar: can you pm me it? or will it block it too? if you dont mind can you write it out by words? idk lol

Reply November 3, 2014 - edited
Avatar

[quote=Valued]yeah i have no clue how to do part b...
can you maybe explain it to me?[/quote]

sorry I tried typing out the formula like 5 times but basil keeps blocking it

Reply November 3, 2014 - edited
Valued

[quote=Avatar]How did you get part a but not b? The questions are basically in reverse,. You had an equation and substituted x and solved for y. Now its giving you y and asking you to solve for x.[/quote]

yeah i have no clue how to do part b...
can you maybe explain it to me?

Reply November 3, 2014 - edited
Avatar

How did you get part a but not b? The questions are basically in reverse,. You had an equation and substituted x and solved for y. Now its giving you y and asking you to solve for x.

too bad I wrote the equation out for you but basil thought I was trying to use to write a shortened link so it deleted what I wrote

Reply November 2, 2014 - edited