how do u solve x^2=13.9 or how to solve Ln(2x-1)^2=4

Note that there are two solutions to x² = 13.9, namely x = +√13.9 and x = -√13.9.

To solve ln(2x-1)² = 4 first use one of the properties of logarithms (see the Formulary) to get

2ln(2x-1) = 4 so that

ln(2x-1) = 2

Now exponentiate:

e^ln(2x-1) = e², so

2x-1 = e², because e^ln(2x-1) is just 2x-1

2x = 1+e²

x = (1+e²)/2

To solve for the x^2 = 13.9 you need to square root both sides.

After square rooting both side you will be left with x= ____ .

For the solving of with the natural log (ln). To remove the ln from the equation you will use the e which removes the Ln from the Left side and leaves (2x-1)^2 and e^4. Now you just need to square everything inside of the parenthesis. Which creates 4x -2 = e^4. After removing the -2 by adding +2 to both sides. Now you are left with 2x = e^4. After plugging in the e^4 you take that answer and divide it by 2 which will leave you with the answer of x=______.

Honestly, I'm questioning your squaring of the pieces inside of the parenthesis. 

Also, I don't think you can square those numbers inside if they are seperated by an operation. I think you would have to FOIL.

However, I could be wrong.