5^3x = 65
3x log5 = log65
3x = log65/log5
x = log65/3log5
I assume you meant
2^(2x+3) = 80
(2x+3)log2 = log80
2x+3 = log80/log2
x = (log80/log2 - 3)/2
Exponetial Functions
In the following exponential relations, solve for x using logarithms. Round your answers to 4 decimal places.
a)
5^3x=65
b)
2^2x+3=80
2 answers
a)
The way you typed it ....
125x = 65
x = 65/125 = 13/25
if you meant: 5^(3x) = 65
take log of both sides, then use log rules
3x log5 = log65
x = log65/(3log5) = appr .865
b) again, the way you typed it...
4x + 3 = 80
4x = 77
x = 77/4
did you mean 2^(2x + 3) = 80 or did you mean
2^(2x) + 3 = 80 ?
Can you see how essential brackets have to be in this type of problem
The way you typed it ....
125x = 65
x = 65/125 = 13/25
if you meant: 5^(3x) = 65
take log of both sides, then use log rules
3x log5 = log65
x = log65/(3log5) = appr .865
b) again, the way you typed it...
4x + 3 = 80
4x = 77
x = 77/4
did you mean 2^(2x + 3) = 80 or did you mean
2^(2x) + 3 = 80 ?
Can you see how essential brackets have to be in this type of problem
2 answers