e^(3x+1) = n
and
log3 (kx) = b
How do you solve for x?
2015-09-02 8:05 am
回答 (3)
2015-09-02 8:31 am
3x+1=ln(n)
x= ...
kx=3^b
x=...
x= ...
kx=3^b
x=...
2015-09-02 12:29 pm
e^(3x+1)=n so
ln(e^(3x+1)=ln(n) so
(3x+1)ln(e)=ln(n) so
3x+1=ln(n) since ln(e)=1,so
x=(ln(n)-1)/3
log3(kx)=b so
kx=3^b so
x=(3^b)/k
ln(e^(3x+1)=ln(n) so
(3x+1)ln(e)=ln(n) so
3x+1=ln(n) since ln(e)=1,so
x=(ln(n)-1)/3
log3(kx)=b so
kx=3^b so
x=(3^b)/k
2015-09-02 8:52 am
e^(3x+1) = n
logₑ[e^(3x+1)] = logₑ(n)
(3x + 1) logₑ(e) = logₑ(n)
3x + 1 = logₑ(n)
3x = logₑ(n) - 1
x = [logₑ(n) - 10 ] / 3
====
log₃(kx) = b
kx = 3^b
x = (3^b) / k
logₑ[e^(3x+1)] = logₑ(n)
(3x + 1) logₑ(e) = logₑ(n)
3x + 1 = logₑ(n)
3x = logₑ(n) - 1
x = [logₑ(n) - 10 ] / 3
====
log₃(kx) = b
kx = 3^b
x = (3^b) / k
收錄日期: 2021-04-20 16:02:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150902000510AAEhMjq