Log R -3 log T + 1/3 Log X
2007-06-25 13:31:39 · 7 answers · asked by roninshinobix 1 in Science & Mathematics ➔ Mathematics
Log R - 3 Log T + (1/3) Log X
power rule of logs: m*log a = log a^m
= Log R - Log T^3 + Log X^(1/3)
= Log (R / T^3) + Log X^(1/3)
= Log [R(X^(1/3)) / T^3 ]
When condensing to a single log term, the arguments of positive log terms are placed in the numerator and the arguments of negative log terms are placed in the denominator.
2007-06-25 13:39:37 · answer #1 · answered by mathjoe 3 · 1⤊ 0⤋
log R - 3 log T + 1/3 log X
= log R - log T^3 + log X^1/3
= log (R X^1/3) / T^3
2007-06-25 13:40:33 · answer #2 · answered by Hell's Angel 3 · 0⤊ 0⤋
Log R -3 log T + 1/3 Log X
log R-log T³+log X^(1/3)
log (rx^1/3)/t³
2007-06-25 13:36:32 · answer #3 · answered by yupchagee 7 · 0⤊ 1⤋
This would be log ((R/t^3)* x^1/3), or log ((r*x^1/3)/t^3)
2007-06-25 13:37:11 · answer #4 · answered by Mark S, JPAA 7 · 0⤊ 0⤋
log R - log T ³ + log X^(1/3)
log R + log X^(1/3) - log T ³
log [ RX^(1/3) / T ³ ]
2007-06-26 08:26:47 · answer #5 · answered by Como 7 · 0⤊ 0⤋
Candido have been given the 1st step yet stopped too quickly. log(80 one/sixteen) = log[80 one] - log[sixteen] = log[3^4] - log[2^4] remember that log[a^b] = b log[a] log(80 one/sixteen) = log[80 one] - log[sixteen] = log[3^4] - log[2^4] = 4 log[3] - 4 log[2] = 4{log[3] - log[2]}
2016-11-07 10:52:51 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Log((R/T^3)*(cube root X)).
2007-06-25 13:40:32 · answer #7 · answered by zee_prime 6 · 0⤊ 0⤋