5 ln x =2 ?
e^-x = 11?
Any idea anybody
2007-02-06 16:19:08 · 7 answers · asked by jeniferrrr 1 in Science & Mathematics ➔ Mathematics
5 ln x = 2
ln x = 2/5
x = e^ 2/5
x = 1.492
e^-x = 11
ln e^-x = ln 11
-x = ln 11
x = -ln 11
x = -2.398
2007-02-06 17:07:47 · answer #1 · answered by M. Abuhelwa 5 · 0⤊ 0⤋
x=e^(2/5)
x= - ln 11
2007-02-07 00:36:03 · answer #2 · answered by Daniel and Fang Fang in China 2 · 0⤊ 0⤋
divide both sides by 5:
ln x =2/5
raise both sides to e
e^(ln x) = e^(2/5)
x = e^(2/5)
e^(-x) = 11
like before but this time take the natural log of both sides
ln[e^(-x)] = ln[11]
-x = ln[11]
x = -ln[11]
2007-02-07 00:24:45 · answer #3 · answered by kellenraid 6 · 0⤊ 0⤋
5 ln x =2 divide both sides by 5
ln x=0.4 take exponential of each side
x=e^0.4=1.4918
e^-x = 11 take ln of each side
-x=ln 11
x=-ln 11=-2.398
2007-02-07 00:29:04 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
1) Whenever you need to invert ln(x) or e^x remember the following theorems
1) If ln(x) = k; then x = e^k
2) if e^x = k, then x = ln(k)
2007-02-07 00:35:39 · answer #5 · answered by 1988_Escort 3 · 0⤊ 0⤋
For the first one:
5lnx = 2; divide both sides by 5
lnx = .4;
e^lnx = e^.4;
x = 1.49
For the second one:
e^-x = 11;
ln(e^-x) = ln11;
-x = 2.40
x = -2.40
2007-02-07 00:32:35 · answer #6 · answered by berman250 2 · 0⤊ 0⤋
How did your teacher or book explain this? Look at your notes or book.
2007-02-07 00:22:40 · answer #7 · answered by smartprimate 3 · 0⤊ 2⤋