SOLUTION: Solve for x: 2log(x+2)=log(3)+log(2x+1) I do not know how to go about solving this problem at all. Also, how do I use superscript in the text box when asking you a ques

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve for x: 2log(x+2)=log(3)+log(2x+1) I do not know how to go about solving this problem at all. Also, how do I use superscript in the text box when asking you a ques      Log On


   

Question 136389: Solve for x:
2log(x+2)=log(3)+log(2x+1)
I do not know how to go about solving this problem at all.

Also, how do I use superscript in the text box when asking you a question? When I type it in a word document and copy/paste it into the text box, all of the superscripts turn back into regular, fully-sized numbers. Thank you.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
So;ve for x:
2log%28%28x%2B2%29%29+=+log%28%283%29%29%2Blog%28%282x%2B1%29%29 Apply the "product rule" to the right side. Log%5Bb%5D%28M%29%2BLog%5Bb%5D%28N%29+=+Log%5Bb%5D%28MN%29
2log%28%28x%2B2%29%29+=+log%28%283%282x%2B1%29%29%29 Apply the "power rule" to the left side.pLog%5Bb%5D%28M%29+=+Log%5Bb%5D%28M%29%5Ep
log%28%28x%2B2%29%29%5E2+=+log%28%283%282x%2B1%29%29%29 therefore:
%28x%2B2%29%5E2+=+3%282x%2B1%29 Simplify.
x%5E2%2B4x%2B4+=+6x%2B3 Subtract (6x+3) from both sides.
x%5E2-2x%2B1+=+0 Factor.
%28x-1%29%28x-1%29+=+0, so the solution is a double root:
x+=+1
Check:
2log%28%28x%2B2%29%29+=+log%28%283%29%29%2Blog%28%282x%2B1%29%29 Substitute x = 1.
2log%28%281%2B2%29%29+=+log%28%283%29%29%2Blog%28%282%281%29%2B1%29%29 Simplify.
2log%28%283%29%29+=+log%28%283%29%29%2Blog%28%283%29%29
2log%28%283%29%29+=+2log%28%283%29%29