sqrt(e^5), sqrt(e^-5) ???
2006-09-29 12:44:59 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
No natural log
2006-09-29 12:46:58 · update #1
x^2=e^5
so
x= e^{5/2} or x= -e^{5/2}
or in your notation:
x=sqrt(e^5), or x=-sqrt(e^5)
notice that the sign of the exponent of e does not change
2006-09-29 12:49:08 · answer #1 · answered by Anonymous · 2⤊ 0⤋
Ln X 5 2
2016-12-16 14:21:31 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Do e to both sides :
e^(ln x^2) = e^5
The left side e "cancels" ln so x^2 = e^5
x = + or - sqrt (e^5)
2006-09-29 12:51:41 · answer #3 · answered by hayharbr 7 · 3⤊ 0⤋
ln(x^2) = 5
x^2 = e^5
x = sqrt(e^5)
or
x = -sqrt(e^5) or sqrt(e^5)
this can also be written as
x = -e^(5/2) or e^(5/2)
2006-09-29 14:12:44 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
ln(x^2) = 5
x^2 = e^5
x = (e^5)^(1/2)
x = e^2.5
hmmmmm, lolitabel...'s right. I forgot the - root.
2006-09-29 12:50:32 · answer #5 · answered by Helmut 7 · 0⤊ 2⤋
lnx² = 5
2ln x = 5
ln x = 5/2
x = ±√(e∙e∙e∙e∙e)
2006-09-29 16:46:39 · answer #6 · answered by Jerry M 3 · 0⤊ 0⤋
ln(x^2) = 5
2 ln(x) = 5
ln(x) = 5/2
x = e^(5/2)
2006-09-29 12:49:31 · answer #7 · answered by language is a virus 6 · 1⤊ 2⤋
well, ln(x^2) = 5
exp(ln(x^2)) = e^5
x^2 = e^5
x= +e^(5/2) or -e^(5/2)
2006-09-29 12:50:38 · answer #8 · answered by spongeworthy_us 6 · 1⤊ 1⤋
x^2=e^5
x=e^2.5=12.18249
2006-09-29 12:52:45 · answer #9 · answered by yupchagee 7 · 0⤊ 2⤋
If by "In" you mean integer, that's impossible.
2006-09-29 12:46:17 · answer #10 · answered by Anonymous · 0⤊ 4⤋