Question

does anyone know how to work -2.9= -log [h^+] particularlly on a calculator?

Answer 1

Still cannot figure out what you're looking for here.

Answer 2

what power is h being raised to?

Answer 3

Let me write the question

Answer 4

if the range ph of an apple is 2.9 to 3.3 use the formula ph=-log [h\[^{+}\] find the difference

Answer 5

close bracket on the plus

Answer 6

find the range

Answer 7

-log (h+ what?

Answer 8

it just has the plus as an exponent

Answer 9

Oh I think I understand..

\[-2.9 = -log(h)\]\[\implies 10^{-2.9} = 10^{-log(h)}\]\[\implies (10^{log(h)})^{-1} = 10^{-2.9}\]\[\implies h^{-1} = 10^{-2.9}\]\[\implies \frac{1}{h} = \frac{1}{10^{2.9}}\]\[\implies h = 10^{2.9} \approx 794.328 \]

Answer 10

Wow, how can this be done on a TI-84 plus?

Answer 11

We have the steps in the book, but we just cannot figure out how to work it on the cal

Answer 12

I did it all by hand except the last step.. You don't need the calculator except to find out what \(10^{2.9}\) is.

Answer 13

The answer is suppose to be 5.01x10 exponent -1 to 1.26x 10 exponent -3 the problem becomes 1.26x10 exponent -3 = [h+] where does the -3 comes from?

Answer 14

Nope, I don't understand any of that. I'm afraid there something wrong with the way the problem is presented.

Answer 15

Hmmm, I will look again. Be right back

Answer 16

ph=-log[h^+]
-2.9=-log [h^+]
-2.9log [h^+]
1.26x10^-3\[\approx[h^+]\]

Answer 17

the range can be from 5.01x10^-4 to 1.26x10^-3

Answer 18

Does this look familar?