How do I solve e^(ax) = Ce^(bx), when a does not equal b?
I know the answer has something to do with natural log, and that a has to be connected to b and c, but I don't know what.
take ln of both sides
ln [e^(ax)] = ln[ C e^(bx) ]
ax lne = lnC + bx lne, using the rules of logs, and recall that lne = 1
ax = lnC + bx
ax - bx = lnC
so when a = b
ax - ax = lnC
0 = ln C
C = 1
when a ≠ b
ax - bx = lnC
x(a-b) = lnC
x = lnC/(a-b)
To solve the equation e^(ax) = Ce^(bx), where a is not equal to b, we can use the properties of logarithms. We will need to take the natural logarithm (ln) of both sides of the equation.
Starting with the equation e^(ax) = Ce^(bx), take the natural logarithm of both sides:
ln(e^(ax)) = ln(Ce^(bx))
Using the logarithmic identity ln(a^b) = b * ln(a), we can rewrite the left side of the equation:
ax ln(e) = ln(Ce^(bx))
Since ln(e) is equal to 1, we have:
ax = ln(Ce^(bx))
Apply the logarithmic property again on the right side, ln(Ce^(bx)):
ax = ln(C) + ln(e^(bx))
As ln(e^(bx)) can be simplified to bx, we have:
ax = ln(C) + bx
Rearranging the equation to isolate x:
ax - bx = ln(C)
Factoring out x:
x(a - b) = ln(C)
Finally, solving for x:
x = ln(C) / (a - b)
Therefore, the solution to the equation e^(ax) = Ce^(bx), when a does not equal b, is x = ln(C) / (a - b).