35= 400e^(-0.4x)?
Using logarithm to the base e,
ln (35) = ln (400) + ln (e^-0.4x)
ln (35) = ln 20^2 + (-0.4x) ln (e); [where ln e = 1]
ln (35) = 2 ln (20) - 0.4x
0.4x = 2 ln (20) - ln (35)
x = [2 ln (20) - ln (35)]/0.4 = 6.0903
Solve for x using logs. Give an exact answer.
35= 400e^(-0.4x)
Could someone point me in the right direction? Applying the logs to e is what's confusing me...
and then finding the inverse function of this problem...
f(t)=77e^(0.2t)
any help is greatly appreciated