SOLUTION: log (2x+1) base 2 - log (x-1) base 2 = 3
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Question 634442: log (2x+1) base 2 - log (x-1) base 2 = 3
Found 3 solutions by lwsshak3, stanbon, MathTherapy:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
log (2x+1) base 2 - log (x-1) base 2 = 3
place under single log base 2
log2[(2x+1)/(x-1)]=3
Convert to exponential form: base(2) raised to log of number(3)=number(2x+1)/(x-1)
2^3=(2x+1)/(x-1)=8
8(x-1)=2x+1
8x-8=2x+1
6x=9
x=9/6=3/2=1.5
..
Check:
log (2*1.5+1) base 2 - log (1.5-1) base 2 = 3
log2(4)-log2(.5)=4-1=3
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
log (2x+1) base 2 - log (x-1) base 2 = 3
-----
log2[(2x+1)(x-1)] = 3
---
(2x+1)(x-1) = 2^3
----
2x^2 -x -1 = 8
--------
2x^2 -x -9 = 0
----
x = [1 +- sqrt(1-4*2*-9)]/4
------
Positive solution:
x = [1+ sqrt(73)]/4
----
x = 2.386...
Cheers,
Stan H.
=============
Answer by MathTherapy(10555) (Show Source): You can put this solution on YOUR website!
log (2x+1) base 2 - log (x-1) base 2 = 3
log[2] (2x + 1) - log[2] (x - 1) = 3
log[2] (2x + 1)/(x - 1) = 3
8(x - 1) = 2x + 1 ----- Cross-multiplying
8x - 8 = 2x + 1
8x - 2x = 1 + 8
6x = 9
x = , or , or
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