Solve the following equation for t

Question

c=a(8−e^bt)

i got:
ln-(c-8a)/a/b.......????

thank you so much

Answer 1

c = a(8-e^bt)
c/a = 8-e^bt
e^bt = 8 - c/a
bt = ln(8 - c/a)
t = ln(8 - c/a)/b

You cannot pull that 1/a out of the log. You could say
t = (ln(8a-c)-lna)/b
I prefer 8a-c to -(c-8a) anyway. I get nervous looking at logs of negative numbers ...