e^(ax) = Ce^(bx)?

0 ⤊

e^(ax) = Ce^(bx)
ln ln
ax = bx * ln (Ce) ???

I'm trying to solve for X.

2007-01-27 16:25:51 · 2 answers · asked by RogerDodger 1 in Science & Mathematics ➔ Mathematics

I meant to type lower case 'e' in the heading (i.e., 2.71828...)

2007-01-27 16:30:42 · update #1

e^(ax) = Ce^(bx)
Log(e^(ax)) = Log(Ce^(bx))
ax = Log(C) + bx
(a-b)x = Log(C)
x = Log(C)/(a-b)

2007-01-27 16:30:05 · answer #1 · answered by Scythian1950 7 · 0⤊ 0⤋

ln both sides
ax = bx + ln C
x = (ln C) / (a - b)

2007-01-27 16:31:09 · answer #2 · answered by kimjay_lmr01 1 · 0⤊ 0⤋