Question

I need help with understanding logarithms, please.
The directions: Find the Value of x.
ln(2x-1)+2=0

Answer 1

\[\ln(2x-1)+2=0\]\[\ln(2x-1)=-2\]\[\ln(2x-1)=-2\times \ln(e)\]\[\ln(2x-1)= \ln(e^{-2})\]\[2x-1=e^{-2}\]\[2x-1=\frac{1}{e^2}\]\[2x=\frac{1}{e^2}+1\]\[2x=\frac{1}{e^2}+\frac{e^2}{e^2}\]\[2x=\frac{e^2+1}{e^2}\]\[x=\frac{e^2+1}{2e^2}\]

Answer 2

you can use 2.7 approximation for e, or leave it the way it is now.

Answer 3

Thankyou SOOO much....that makes a lot of sense :)

Answer 4

You welcome:) Glad to help !

Answer 5

Fixed your color :)