# Thread: Solving ln sqrt(x+4) = 1

1. ## Solving ln sqrt(x+4) = 1

just one more and im done

so ln sqrt(x+4) = 1

= e^1 = sqrt(x+4)...square both sides

e^2 = x+4
e^2-4 = x

now I need to solve by hand in exact form. Is this as far as I can go?

2. ## Re: quick question on product rule and equality rule for logs

Originally Posted by RYdis
just one more and im done

so ln sqrt(x+4) = 1

= e^1 = sqrt(x+4)...square both sides Get rid of that first equals sign

e^2 = x+4
e^2-4 = x

now I need to solve by hand in exact form. Is this as far as I can go?
Yes, this is the answer in exact form. The only thing you might be able to do is factorise the LHS using Difference of Two Squares...

3. ## Re: quick question on product rule and equality rule for logs

Originally Posted by Prove It
Yes, this is the answer in exact form. The only thing you might be able to do is factorise the LHS using Difference of Two Squares...
(e-2)(e+2) = x?

since squares can't be negative..does this leave just e+2 = x instead of
e+2 = x or e-2 = x. should I just leave as the difference?

4. ## Re: quick question on product rule and equality rule for logs

Originally Posted by RYdis
(e-2)(e+2) = x?

since squares can't be negative..does this leave just e+2 = x instead of
e+2 = x or e-2 = x. should I just leave as the difference?
No, x = (e-2)(e+2). You can't go any further.

5. ## Re: quick question on product rule and equality rule for logs

Originally Posted by Prove It
No, x = (e-2)(e+2). You can't go any further.
awesome, thanks for help, starting to get it now.