Solve for t using the power rule and other rules

Question

1. e^-.3=64

2 e^kt=1/3

3. e^(ln0.6)t=0.9

I have tried them out and no luck. Any help with these?

Answer 1

For e (the base of the natural logarithm equal to 2.718281828), e^-0.3t = 64

is rewritten as -0.3t × ln e = ln 64 or -0.3t × 1 = ln 64. Then t is ln 64 / -0.3

or -13.86294361.

For e^kt = 1/3, develop kt × ln 2.718281828 = ln (1/3).

This goes to kt × 1 = ln (1/3) or kt = -1.098612289 and

t is -1.098612289 divided by k.

For e^{(ln 0.6)t} = 0.9, restate as [e^{(ln 0.6)}]^t = 0.9.

Then [e^{(ln 0.6)}]^t = 0.9 translates as 0.6^t = 0.9.

This last gives t × ln 0.6 = ln 0.9 and t is

(ln 0.9 ÷ ln 0.6) or 0.2062553458.