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).