e^ax = Ce^bx, where a ≠ b
how do i solve for x?
taking natural log ... ax = bx + ln(C)
ax - bx = ln(C) ... x (a - b) = ln(C)
x = [ln(C)] / (a - b)
e^ax / e^bx = C
e^((a-b)x) = C
(a-b)x = lnC
x = lnC/(a-b)
thank you!
To solve for x in the equation e^ax = Ce^bx, where a ≠ b, we can take the natural logarithm (ln) of both sides of the equation. The natural logarithm is the inverse of the exponential function.
Taking the natural logarithm of both sides:
ln(e^ax) = ln(Ce^bx)
Using the property of logarithms that ln(a^b) = b * ln(a):
ax * ln(e) = ln(C) + bx * ln(e)
The natural logarithm of e is 1, so this simplifies to:
ax = ln(C) + bx
We can rearrange this equation to solve for x:
ax - bx = ln(C)
Factoring out x:
x(a - b) = ln(C)
Finally, dividing both sides of the equation by (a - b):
x = ln(C) / (a - b)
Therefore, the value of x can be found by calculating the natural logarithm of the constant C and dividing it by the difference between a and b.