I really need help. The first one:
6e^-0.07x = e^0.02x
The second one:
30(1.05)^x = 15e^x
Easy 10 points!
2007-12-02 11:54:50 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Take the ln of both sides and apply the rules of logs:
ln 6e^(-0.07x) = ln e^0.02x
ln 6 + -0.07x(ln e) = 0.02x(ln e)
ln 6 = x(0.07 + 0.02)
ln 6 = (0.09)x
x = ln 6 / 0.09
approx. 19.9
second one works the same way:
ln 30 + x ln 1.05 = ln 15 + x ln e
x ln 1.05 - x = ln 15 - ln 30
x (ln 1.05 - 1) = ln (15/30)
x = ln 0.5 / (ln 1.05 - 1)
that's it! :)
2007-12-02 12:07:37 · answer #1 · answered by Marley K 7 · 0⤊ 0⤋
6e^-0.07x = e^0.02x
Take ln,
ln6 - 0.07x = 0.02x
x = ln6/0.09
30(1.05)^x = 15e^x
Take ln,
ln30 + xln1.05 = ln15 + x
x = ln2/(1-ln1.05)
2007-12-02 20:00:29 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
6e^(-0.07X) = e^(0.02X)
ln6 - 0.07X = 0.02X
ln6 = 0.09X
X=ln6/0.09
30(1.05)^X=15e^X
ln30 + X*ln1.05 = ln15 + X
solve from there...
2007-12-02 20:03:43 · answer #3 · answered by Trouser 3 · 0⤊ 0⤋
6 e^-.07x = e^.02x
6 = e^.02x / e^-.07x
6 = e^(.02+.07)x = e^.09x
ln(6) = .09x
ln(6) / .09 = x
30(1.05)^x = 15e^x
2 (1.05)^x = e^x
ln ( 2 ( 1.05)^x ) = x
ln (2) + ln ( (1.05)^x ) = x
ln (2) + x ln (1.05) = x
ln (2) = x ( 1 - ln (1.05))
ln (2) / ( 1 - ln(1.05)) = x
2007-12-02 20:01:48 · answer #4 · answered by jgoulden 7 · 0⤊ 0⤋