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Tutors Answer Your Questions about logarithm (FREE)
Question 94815: I am currently enrolled in a General Chemistry course at a community college. I should apparently already know logarithms, but I don't. Please help!
The problem is as follows...
The pK_a_ of a solution is defined by the equation: pK_a_ = -logK_a_
Where: K_a_ = acid dissociation constant
Use the rules for logarithms and exponents to solve for K_a_ in terms of pK_a_.
(a.) K_a_ = _____________
(b.) If pK_a_ = 1.64 then K_a_ = ________________
I have been able to solve problem a. K_a_ = 10^-pK_a_^
However, I cannot figure out how to get the answer to problem b.
I know that the correct answer is 2.29E-2 but I can't figure out how they get that answer.
Any help would be greatly appreciated!
Thank you!
Amy M
a.mccubbin@insightbb.com
Click here to see answer by scott8148(3382)   |
Question 95610: dear tutors i have been stumped with this question that did not come from a textbook, please try to answer it for me, thanks.
log(2x) + log(x-1) - 2 log(x) = 1. you need to solve for x. i tried converting it into exponential form 10 = 2x(x-1)/x^2, = (2x^2-2x)/x^2 = (2x-2)/x and then multiplication of both sides by x to yield: 10x = 2x-2, then subtraction of 2x from both sides becomes: 8x = -2, division of both sides by 8 yields: x = -1/4 but it is impossible for x to have a negative value since it is impossible to perform a logarithm on a negative? thank you so much for taking the time to figure this out i really appreciate it
Click here to see answer by stanbon(26252)  |
Question 96003: my problem is:
log (base 4) 9x^2 = log (base 2)(5x-8)
i think i have to switch the base on the left hand side to 2 which would give me
log (base 2) 9x^2
----------------- = log (base 2)(5x-8)
log (base 2) 4
i dont even know if thats right..but after that im completely lost though...any help would be appreciated, thanks.
Click here to see answer by scott8148(3382) |
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