SOLUTION: Q: e^(2x+5)=40
its pretty simple i think, this is what i've figured out.
ln(e^(2x+5))=ln(40)
2x+5=ln(40)
2x+5 = 3.688
2x = -2.688
x = -1.344
im not sure if thats right,
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: Q: e^(2x+5)=40
its pretty simple i think, this is what i've figured out.
ln(e^(2x+5))=ln(40)
2x+5=ln(40)
2x+5 = 3.688
2x = -2.688
x = -1.344
im not sure if thats right,
Log On
Question 152520: Q: e^(2x+5)=40
its pretty simple i think, this is what i've figured out.
ln(e^(2x+5))=ln(40)
2x+5=ln(40)
2x+5 = 3.688
2x = -2.688
x = -1.344
im not sure if thats right, i didnt really pay attention in logs,and e's. Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! Quickest way to check is to plug your solution into the equation and make sure it solves it.
Oops, there's a problem.
You paid attention for logarithms but not during subtraction.
Your thinking and approach to the problem are clear.
It should be,
Checking,
Close enough.
