2006-07-26 16:11:02 · 6 answers · asked by ashka e 1 in Science & Mathematics ➔ Mathematics
log(7)m = (1/3)log(7)64 + (1/2)log(7)121
log(7)m = log(7)(64^(1/3)) + log(7)(121^(1/2))
log(7)m = log(7)4 + log(7)11
log(7)m = log(7)(4 * 11)
log(7)m = log(7)44
m = 44
2006-07-27 03:41:00 · answer #1 · answered by Sherman81 6 · 2⤊ 0⤋
4
2006-07-26 16:14:26 · answer #2 · answered by Danny H 3 · 0⤊ 0⤋
I strongly suspect your problem is log7^m=1/3log7^64+1/2log7^121, or m * log7 = 1/3log7^64+1/2log7^121. Otherwise, it is way too complicated. Let's assume the first one is the correction equation for now.
log7^m=1/3log7^64+1/2log7^121
= log7^(64/3) + log7^(121/2)
= log7^(64/3 + 121/2)
=log7^(491/6)
Then, m = 491/6.
Note that since log7^m = m * log7, if thee right question is the second one, the anser is the same. m=491/6.
2006-07-26 20:16:56 · answer #3 · answered by Stanyan 3 · 0⤊ 0⤋
I think you mean
log_7 m = 1/3 log_7 64 + 1/2 log_7 121
and then solve for m.
Remember to use these identities:
n log_b x = log_b x^n
log_b m + log_b n = log_b mn
Therefore, (using one of our identities)
log_7 m = log_7 64^(1/3) + log_7 121^(1/2)
Simplifying the exponents:
log_7 m = log_7 4 + log_7 11
simplify the right side(using one of our identities)
log_7 m = log_7 4(11)
Therefore,
m = 4(11)
m = 44
^_^
2006-07-27 01:07:34 · answer #4 · answered by kevin! 5 · 0⤊ 0⤋
if your solving for m,
log7m = 491/6 (log 7)
if you solve this m = 2.05 e 68.
The guy above me is most probably right because you rarely see a question that would yield such a big number for an answer
2006-07-26 20:18:51 · answer #5 · answered by aking 2 · 0⤊ 0⤋
Is log L-O-G or the word log?
2006-07-26 16:26:10 · answer #6 · answered by Asterisk_Love♥ 4 · 0⤊ 0⤋