you want to solve for t
60000 e^.0675t = 80000
e^.0675t = 1.333...
ln both sides and use rules of logs
.0675t lne = ln 1.3333... ( but lne = 1)
t = ln 1.3333../.0675 = 4.26 years, appr 4 years and 3 months
(1/8)^(2x-3) = 16^(x+1)
(2^-3)^(2x-3) = (2^4)^(x+1)
2^(-6x + 9 = 2^(4x + 4)
-6x + 9 = 4x + 4
-10x = -5
x = 1/2
e^(4x) = 4^(x-2)
ln both sides
4x lne = (x-2)ln4
4x = ln4 x - 2ln4
4x - ln4 x = -2ln4
x(4 - ln4) = -2ln4
x = -2ln4/(4-ln4)
= appr -1.0608
15. ln(x-4) ln3 = ln(1/5)^x
no simple algebraic way to solve this, unless I misread the question. I took it the way you typed it
ran it through Wolfram and Wolfram ran out of time trying to solve it
1 e^.068t = 3
now do it the same way as #19
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