2007-09-02 17:25:28 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
3logu = log(u^3)
(1/2)logz = log(sqrt(z))
log a + log b - log c = log(ab/c)
3log u + log v -1/2log z
= log(u^3 * v / sqrt(z))
2007-09-02 17:30:14 · answer #1 · answered by gudspeling 7 · 0⤊ 0⤋
the logarithm power rule or whatever its called so it'll be log x^1/2 + log y - log z^2 . then you use the adding subtraction logarithm rule. log [(x^1/2 y )/ (z^2)]
2016-05-19 23:08:31 · answer #2 · answered by Anonymous · 0⤊ 0⤋
3log u + log v -1/2log z
= log u^3 + log v - log z^1/2
= log (u^3v/z^1/2)
2007-09-02 17:31:31 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
log(vu^3/z^(1/2))
log rules say that if you are adding two logs than you can combine with multiplication...and if subtracting, combine with division. You can also move any number in front of a log up to the power.
2007-09-02 17:30:20 · answer #4 · answered by eman 2 · 0⤊ 0⤋
3 log(u) + log v - 1/2 log z =
log(u)^3 + log v - log (z)^1/2 (since a log x = log x^a)
log(u)^3(v) - log(z)^1/2 ( since log a + log b = log(ab))
log[(u)^3 (v)/log(z)^1/2] ( since log a - log b = log a/b)
2007-09-02 17:33:15 · answer #5 · answered by mohanrao d 7 · 0⤊ 0⤋
3logu = log u^3
logv = log v
-1/2 log z = log (1/(z^1/2)) [or log(1/sqrt(z))
---------
sum:
log (u^3 * v / sqrt(z))
2007-09-02 17:31:57 · answer #6 · answered by Anonymous · 0⤊ 0⤋