how do you evaluate...
ln (1/e)^9
and
log(.01)^(1/3)
2006-11-29 05:11:12 · 8 answers · asked by puffer fish 5 in Science & Mathematics ➔ Mathematics
In the first, 1/e = e^-1, and (e^-1)^9 = e^-9. Since ln x is the inverse of e^x, ln(e^-9) = -9
In the 2nd, .01 = 10^-2, so log[ (10^-2)^(1/3)]= log 10^(-2/3) = -2/3
2006-11-29 05:21:00 · answer #1 · answered by Philo 7 · 0⤊ 0⤋
first one :
ln ((1/e)^9) = 9 *ln (1/e) = 9 * (ln 1 - ln e) = 9 * ( 0 -1) = -9
the second one (if log stands for a base 10 log )
log ((.01) ^(1/3)) = 1/3 *log(10^ (-2)) = (-2/3) * log 10 = -2/3
(tip : review your log functions properties. See if this link can help)
2006-11-29 05:25:47 · answer #2 · answered by Anonymous · 0⤊ 0⤋
By simple rules of logaritms:
ln(1/e)^9 = 9*[ln(1) - ln(e)] = 9*[ 0 - 1] = - 9.
And log(.01)^(1/3) = (1/3)*[log(.01)] = (1/3)*[log(1/100)] = (1/3)*
[log(1) - log(100)] = (1/3)*[0 - 2] = - 2/3
2006-11-29 05:21:49 · answer #3 · answered by quidwai 4 · 0⤊ 0⤋
you can take the power 9 out to the front:
9*ln(1/e)
then split it up using the logarithm rules:
9*(ln(1)-ln(e))
ln(1)=0 and ln(e)=1 so
9*(0-1)=-9
do something similar with the second:
(1/3)*log(1/100)=(1/3)*(log(1)-log(10^2))
=(1/3)*(0-2*log(10))
log(10)=1, so you have this:
(1/3)*(-2)=-2/3
2006-11-29 05:22:18 · answer #4 · answered by nemahknatut88 2 · 0⤊ 0⤋
For the first one calculate ln(1/e) and then take the result to the 9th power.
The second one take the log (assuming base 10) of 0.1 and then find the cube root of the result.
2006-11-29 05:19:38 · answer #5 · answered by Anonymous · 0⤊ 0⤋
f(x)=log(base 2)x at x=a million/4 f(a million/4) = log(base 2) a million/4 = m 2^m = a million/4 -----> 2^m = a million/(2^2) -----> 2^m = 2^(-2) -----> m = -2 f(a million/4) = -2 that's your answer. in case you probably did no longer particularly get what I did with a million/4, look: a million/4 = a million/(2^2) regardless of the undeniable fact that, a million = 2^0 so we are able to rewrite this as: = (2^0) / (2^2) that's a branch of powers with an identical base so we subtract the exponents: = 2^(0 - 2) = 2^(-2) i desire this helps...
2016-10-13 08:56:13 · answer #6 · answered by Anonymous · 0⤊ 0⤋
ln(1/e)^9 = 9ln(1/e) = 9[ln(1) - ln(e)] = 9[0 - 1] = -9
log(.01)^1/3 = 1/3log(10^(-2)) = (1/3)(-2)log10 = -2/3
2006-11-29 05:21:00 · answer #7 · answered by kellenraid 6 · 0⤊ 0⤋
ln(1/e)^9 =
9*[ln(1/e)] = 9*[ln(1) - ln(e)] = 9*[ 0 - 1] = -9
*************************************************
log(0.01)^(1/3) =
(1/3)*log(0.01) = -2/3
since y = log(0.01) iff 10^y = 0.01, then we know y = -2
2006-11-29 05:26:17 · answer #8 · answered by Anonymous · 0⤊ 0⤋