# Solve Logarithms symbolically

• Jun 15th 2008, 03:58 PM
AndyHirte
Solve Logarithms symbolically
Ok, this is my first post so here goes.

I am trying to solve
Use common or natural logarithms to solve symbolically.

2e^2x+2= 6
first I divide both sides by 2
e^2x+2= 6
use natural logarithms
ln e^2x+2 = ln 6
Simplify
2x+2 = ln 6
subtract 2 from both sides
2x = ln 6-4
Divide by 2
x = ln 6-4/2

I enter this into my Ti-83plus
ln (6-4)/2
my answer is .34657 It is wrong I figure it is the way I have entered it into the calc. Anybody have any thoughts of what is wrong?
• Jun 15th 2008, 04:04 PM
mr fantastic
Quote:

Originally Posted by AndyHirte
Ok, this is my first post so here goes.

I am trying to solve
Use common or natural logarithms to solve symbolically.

2e^2x+2= 6
first I divide both sides by 2
e^2x+2= 6 Mr F says: This line is wrong. You have to divide all terms by 2, not just the first ....... e^2x + 1 = 3. Personally, my first step would be to subtract 2 from both sides: 2 e^2x = 4 .......

use natural logarithms
ln e^2x+2 = ln 6
Simplify
2x+2 = ln 6
subtract 2 from both sides
2x = ln 6-4
Divide by 2
x = ln 6-4/2

I enter this into my Ti-83plus
ln (6-4)/2
my answer is .34657 It is wrong I figure it is the way I have entered it into the calc. Anybody have any thoughts of what is wrong?

..
• Jun 15th 2008, 04:37 PM
AndyHirte
Thank you for the pointers, still not getting the right answer

e^2x+1 = 3
ln e^2x+1 = ln 3
2x+1 = ln 3
2x = ln 3-1
x = ln 3-1/2
.34657
still not the right answer. I am doing something else wrong?
• Jun 15th 2008, 04:40 PM
mr fantastic
Quote:

Originally Posted by AndyHirte
Thank you for the pointers, still not getting the right answer

e^2x+1 = 3
ln e^2x+1 = ln 3 Mr F says: No. That's totally invalid. Subtract 1 from both sides: e^2x = 2. Now take the log of both sides.

2x+1 = ln 3
2x = ln 3-1
x = ln 3-1/2
.34657
still not the right answer. I am doing something else wrong?

..