express logз (392)^1/3 in terms of h and k
2006-08-04 05:37:11 · 11 answers · asked by Daphne Loo 1 in Science & Mathematics ➔ Mathematics
Given ㏒3 5= h and ㏒3 7= k
express㏒3 ³√392 in terms of h and k
2006-08-04 06:26:32 · update #1
Well, I'm not sure how the "logз 5 = h" helps, but here's what I was able to put together:
logз(392^(1/3)) = logз[(2³7²)^(1/3)]
= logз[2*7^(2/3)]
= logз2 + logз7^(2/3)
= logз2 + (2/3)logз7
= logз2 + (2/3)k
Hope that helps!
2006-08-04 06:23:08 · answer #1 · answered by Jay H 5 · 1⤊ 0⤋
Since 2 = 7-5 = 3^k-3^h, an alternate form of the answer is
log3(3^k-3^h) + 2/3k
2006-08-04 09:36:19 · answer #2 · answered by oldbutcrafty 2 · 0⤊ 0⤋
Following Jay H's comment, can you use the approximation that (7^2)/(5^2) ~2 here?
2006-08-04 07:04:59 · answer #3 · answered by Stephan B 5 · 0⤊ 0⤋
5 = 3^h 7 = 3^k
I can reduce it to this:
log (base3) 2 + 2/3 log (base 3) 7
which is the same as:
log (base3) 2 + 2/3k
=> log (base3)10 - h + 2/3k
Sorry, I think this is about as far as you can go.
2006-08-04 06:26:05 · answer #4 · answered by Anonymous · 0⤊ 0⤋
log(3)5 = h
log(3)7 = k
log(3)(392^1/3)
log(3)(7^2 * 2^3)
log(3)(7^2) + log(3)(2^3)
2log(3)(7) + 3log(3)(2)
if by log(3)5, you really mean log(3)2, then
3log(3)h + 2log(3)k
2006-08-04 16:31:06 · answer #5 · answered by Sherman81 6 · 0⤊ 0⤋
Tom nailed it.
(2/3)k + log(base3)(10) - h
2006-08-04 07:24:08 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Express what in terms of what what?
2006-08-04 06:37:48 · answer #7 · answered by Henry 5 · 0⤊ 0⤋
A little bored?? ok maybe more than just a little huh!!
2006-08-04 05:42:36 · answer #8 · answered by A O 2 · 0⤊ 0⤋
what ever
2006-08-04 23:28:47 · answer #9 · answered by Anonymous · 0⤊ 0⤋
my calculator is broken.....
2006-08-04 05:44:35 · answer #10 · answered by paulrb8 7 · 0⤊ 0⤋