SOLUTION: Hi, this is my second time asking a question....both were on logarithms :( Solve for x. {{{ log( 2, x+1 ) }}} = {{{ log( 4, x+3 ) }}} Because {{{ log( 4, x+3 ) }}} can be re

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hi, this is my second time asking a question....both were on logarithms :( Solve for x. {{{ log( 2, x+1 ) }}} = {{{ log( 4, x+3 ) }}} Because {{{ log( 4, x+3 ) }}} can be re      Log On


   

Question 891981: Hi, this is my second time asking a question....both were on logarithms :(
Solve for x.
+log%28+2%2C+x%2B1+%29+ = +log%28+4%2C+x%2B3+%29+
Because +log%28+4%2C+x%2B3+%29+ can be rewritten as +%281%2F2%29log%28+2%2C+x%2B3+%29+,
it was +log%28+2%2C+x%2B1+%29+ = +%281%2F2%29log%28+2%2C+x%2B3+%29+. Then,
+log%28+2%2C+x%2B1+%29+ = +log%28+2%2C+%28x%2B3%29%5E%281%2F2%29+%29+
Thus, +log%28+2%2C+%28x%2B1%29%2Fsqrt%28x%2B3%29+%29+
But I don't know what to do next. Or did I do wrong from the beginning?
I know there could be another way... for example... making +log%28+4%2C+x%2B3+%29+ = +log%28+10%2C+x%2B3+%29%2Flog%28+10%2C+4+%29+ to make the same base for both sides. But still, I don't know what to do next.
Could you please teach me how to solve this problem in those two ways?
Found 2 solutions by Alan3354, MathTherapy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x.
+log%28+2%2C+x%2B1+%29+ = +log%28+4%2C+x%2B3+%29+
-------------
+log%284%2Cx%2B3%29 can be rewritten as %281%2F2%29log%282%2Cx%2B3%29
--------------
log%282%2Cx%2B1%29 = log%284%2Cx%2B3%29%2F2
2log%282%2Cx%2B1%29 = log%282%2C%28x%2B3%29%29
%28x%2B1%29%5E2+=+x%2B3+=+x%5E2+%2B+2x+%2B+1
x%5E2+%2B+x+-+2+=+0
(x-1)*(x+2) = 0
x = 1
=============
x = -2 is rejected, gives log of a neg value

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, this is my second time asking a question....both were on logarithms :(
Solve for x.
+log%28+2%2C+x%2B1+%29+ = +log%28+4%2C+x%2B3+%29+
Because +log%28+4%2C+x%2B3+%29+ can be rewritten as +%281%2F2%29log%28+2%2C+x%2B3+%29+,
it was +log%28+2%2C+x%2B1+%29+ = +%281%2F2%29log%28+2%2C+x%2B3+%29+. Then,
+log%28+2%2C+x%2B1+%29+ = +log%28+2%2C+%28x%2B3%29%5E%281%2F2%29+%29+
Thus, +log%28+2%2C+%28x%2B1%29%2Fsqrt%28x%2B3%29+%29+
But I don't know what to do next. Or did I do wrong from the beginning?
I know there could be another way... for example... making +log%28+4%2C+x%2B3+%29+ = +log%28+10%2C+x%2B3+%29%2Flog%28+10%2C+4+%29+ to make the same base for both sides. But still, I don't know what to do next.
Could you please teach me how to solve this problem in those two ways?


log+%282%2C+x+%2B+1%29+=+log+%284%2C+x+%2B+3%29

log+%282%2C+x+%2B+1%29+=+log+%28%28x+%2B+3%29%29%2Flog+4 ------ Applying change of base

log+%282%2C+x+%2B+1%29+=+%28log+%282%2C+%28x+%2B+3%29%29%29%2Flog+%282%2C+4%29 ----- Changing base on right to base 2

log+%282%2C+x+%2B+1%29+=+%28log+%282%2C+x+%2B+3%29%29%2F2

2log+%282%2C+%28x+%2B+1%29%29+=+log+%282%2C+x+%2B+3%29 ------- Cross-multiplying

log+%282%2C+%28x+%2B+1%29%5E2%29+=+log+%282%2C+x+%2B+3%29

%28x+%2B+1%29%5E2+=+x+%2B+3

x%5E2+%2B+2x+%2B+1+=+x+%2B+3

x%5E2+%2B+2x+-+x+%2B+1+-+3+=+0

x%5E2+%2B+x+-+2+=+0

(x - 1)(x + 2) = 0

highlight_green%28highlight_green%28x+=+1%29%29	   	      OR	         x = - 2 (ignore)



After you'd obtained, +log%28+2%2C+x%2B1+%29+ = +log%28+2%2C+%28x%2B3%29%5E%281%2F2%29+%29+, you should have had: x+%2B+1+=+%28x+%2B+3%29%5E%281%2F2%29, which is

the same as: x+%2B+1+=+sqrt%28x+%2B+3%29

%28x+%2B+1%29%5E2+=+x+%2B+3 ------- Squaring both sides

Go on to solve for x.