SOLUTION: log2 (2x-1)=log4 (3x^2-4x+2)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: log2 (2x-1)=log4 (3x^2-4x+2)      Log On


   

Question 991939: log2 (2x-1)=log4 (3x^2-4x+2)
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

log%282+%2C%282x-1%29%29=log%284+%2C%283x%5E2-4x%2B2%29%29......change to base ten
log%28%282x-1%29%29%2Flog%282+%29=log%28%283x%5E2-4x%2B2%29%29%2Flog%284%29+
log%28%282x-1%29%29%2Flog%282+%29=log%28%283x%5E2-4x%2B2%29%29%2Flog%282%5E2%29+
log%28%282x-1%29%29%2Flog%282+%29=log%28%283x%5E2-4x%2B2%29%29%2F2log%282%29+....cross multiply

2%2Alog%28%282x-1%29%29=log%28%283x%5E2-4x%2B2%29%29+
log%28%282x-1%29%5E2%29=log%28%283x%5E2-4x%2B2%29%29+........if log same, we have
%282x-1%29%5E2=3x%5E2-4x%2B2+......solve for x
4x%5E2-4x%2B1=3x%5E2-4x%2B2+
4x%5E2-3x%5E2-4x%2B4x%2B1-2=0+
x%5E2-1=0+
%28x-1%29%28x%2B1%29=0+
solutions:
%28x-1%29=0+=>x=1
%28x%2B1%29=0+=>x=-1....disregard negative solution,
so, your solution is highlight%28x=1%29



Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

log2 (2x-1)=log4 (3x^2-4x+2)
log+%282%2C+%282x+-+1%29%29+=+log+%284%2C+%283x%5E2+-+4x+%2B+2%29%29

log+%282%2C+%282x+-+1%29%29+=+%28log+%28%283x%5E2+-+4x+%2B+2%29%29%29%2Flog+4 ----- Changing base on right side to base 10

log+%282%2C+%282x+-+1%29%29+=+log+%282%2C+%283x%5E2+-+4x+%2B+2%29%29%2Flog+%282%2C+4%29 ------- Changing base on right side to base 2

log+%282%2C+%282x+-+1%29%29+=+log+%282%2C+%283x%5E2+-+4x+%2B+2%29%29%2F2

2+%2A+log+%282%2C+%282x+-+1%29%29+=+log+%282%2C+%283x%5E2+-+4x+%2B+2%29%29 ----- Cross-multiplying

log+%282%2C+%282x+-+1%29%5E2%29+=+log+%282%2C+%283x%5E2+-+4x+%2B+2%29%29

%282x+-+1%29%5E2+=+3x%5E2+-+4x+%2B+2

4x%5E2+-+4x+%2B+1+=+3x%5E2+-+4x+%2B+2

4x%5E2+-+3x%5E2+-+4x+%2B+4x+%2B+1+-+2+=+0

x%5E2+-+1+=+0

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

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