Simplify this equation? and finding the maximum

e^t = 3e^(t/2)

I know you need to take the natural log of each side, but i don't know how to do that for the right hand side.

The original equation is y = -e^t + 3e^(t/2)
How do i find the maximum of this?

Simplify e^t = 3e^(t/2) by applying natural log, denoted by ln

ln(e^t) = ln(3e^(t/2))
t = ln(3) + ln(e^(t/2))
t = ln(3) + t/2
t - t/2 = ln(3)
t/2 = ln(3)
t = 2 ln(3)
t = ln(9).

In order to find the maximum of y = -e^t + 3e^(t/2), we take its derivative and set it equals zero.

y = -e^t + 3e^(t/2)
dy/dt = -e^t + (3/2)e^(t/2)

Make it equals zero

-e^t + (3/2)e^(t/2) = 0
(3/2)e^(t/2) = e^t

Take the natural log of each side to eliminate the exponencial form

ln((3/2)e^(t/2)) = ln(e^t)

Using properties of log

ln(3/2) + t/2 = t
t - t/2 = ln(3/2)
t/2 = ln(3/2)
t = 2 ln(3/2)
t = ln(9/4)

The function y = -e^t + 3e^(t/2) has a global maximum at t = ln(9/4).

y = -e^(t) + 3e^(t/2)

Differentiate with respect to t:

dy/dt = -e^(t) + (3/2)e^t/2)

At a maximum dy/dt = 0. Hence:

-e^(t) + (3/2)e^(t/2) = 0

=> (3/2)e^(t/2) = e^(t)

=> 3e^(t/2) = 2e^(t)

Take log both sides:

=> ln[(3e^(t/2)] = ln [2e^(t)]

By log laws:

=> ln(3) + ln(e^t/2) = ln(2) + ln(e^t)

By log index law:

=> ln(3) + (t/2)ln(e) = ln(2) + t ln(e)

=> ln(3) + t/2 = ln(2) + t since ln(e) = 1

=> ln(3) - ln(2) = t - (t/2)

=> t/2 = ln(3) - ln(2)

=> t = 2[ln(3) - ln(2)]

=> t = ln(9) - ln(4)

=> t = ln(9/4) exact answer

Evaluate with a calculator

=> t = 0.811 (3 sig.figs)

Ans: max at t = ln(9/4) or t = 0.811 (3 s.f)

first of all you'll get -e^t + (3/2)e^(t/2) upon differentiating the given equation.
so e^t = 3/2 e^(t/2)
taking log
t= log(3/2) + t/2
solve..

e^t/e^(t/2)=3=e^(t/2)

(t/2)ln e=ln 3

t/2=ln 3
t=2 ln 3=ln 3^2=ln 9

God bless you.