2007-12-24 06:38:01 · 7 answers · asked by sparkle_disliker 3 in Science & Mathematics ➔ Mathematics
Method 1
ln 2 + ln e^(5x) = 4
5x ln e = 4 - ln 2
5x = (4 - ln 2)
x = (1/5) (4 - ln 2)
Method 2
2e^(5x) = e^4
e^(5x) = (1/2) e^4
5x = ln(1/2) + 4
5x = ln1 - ln2 + 4
5x = 4 - ln 2
x = (1/5) (4 - ln 2)
2007-12-24 07:18:16 · answer #1 · answered by Como 7 · 3⤊ 0⤋
2e^5x = e^4
4 = ln(2) +5x , get x
2007-12-24 14:46:51 · answer #2 · answered by Nur S 4 · 1⤊ 0⤋
ln(2 e^(5x)) = 4
2 e^(5x) = e^4
e^(5x) = e^4 / 2
5x = ln(e^4 / 2)
x = ln(e^4 / 2) / 5
x = (ln(e^4) - ln(2)) / 5
x = (4 - ln(2)) / 5
This is because the exponential function is the inverse of the natural logarithm.
This identity is also used: log(cd) = log(c) + log(d)
2007-12-24 14:48:19 · answer #3 · answered by Anonymous · 1⤊ 0⤋
ln (2e^(5x)) = 4
ln 2 + ln e^(5x)) = ln 4
ln e^(5x) = ln 4 - ln 2 = ln 4/2 = ln 2 = 0.69314
5x = .69314
x = .13862
2007-12-24 18:54:56 · answer #4 · answered by Joe L 5 · 0⤊ 2⤋
ln(2e^5x)=4
ln2 + lne^5x = 4
ln2 + 5xlne = 4
5x = 4 - ln2
x = 4 - ln2 / 5
x=0.6613705639
2007-12-24 15:26:07 · answer #5 · answered by Murtaza 6 · 0⤊ 2⤋
ln[2exp(5x)]=4
write as:
ln[2]+ln[exp(5x)]=4
ln[2]+5x=4
5x=4-ln[2]
x=1/5(4-ln[2])=0.6614
2007-12-24 14:43:19 · answer #6 · answered by kuiperbelt2003 7 · 1⤊ 0⤋
ln(2e^(5x))=4
ln(2)+ln(e^(5x))=4
ln(2)+5x=4
x=(4-ln(2))/5
2007-12-24 14:41:49 · answer #7 · answered by zazensoto 3 · 2⤊ 0⤋