SOLUTION: Solve: logbase2(1-x)+logbase2(4x+1)=0

Question 130175: Solve: logbase2(1-x)+logbase2(4x+1)=0
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve: logbase2(1-x)+logbase2(4x+1) = 0
log2[(1-x)(4x+1)] = 0
(1-x)(4x+1) = 2^0
-4x^2+3x+1 = 1
4x^2-3x = 0
x(4x-3)=0
x = 0 or x = 3/4
Checking:
If x = 0:
log2(1)+log2)1) = 0
True
--------------
If x = 3/4
log2(1-(3/4)) + log2(4(3/4)+1) = 0
log2(1/4) + log2(4) = 0
-2 + 2 = 0
True
===========
Cheers,
Stan H.
RELATED QUESTIONS
Solve: a) log(x-2)+log(x+1)=1 b) 4(^1-x)=3*5^x c)... (answered by edjones)

solve the simultaneous equation 2(logbase2 tox-1)=logbase2 to y ......eq1 y=x-1 (answered by Alan3354)

Solve for "x" logbase2 4 + logbase2 27= logbase2... (answered by Alan3354,nerdybill)

SOLVE for x: a) logbase3 x + logbase3 (x-8)=2 b) logbase4 (x+2)+logbase4(x-4)=2 c)... (answered by edjones)

Solve with exact solutions log2(x)-log2(3x-1)=0 I only got this far... (answered by oscargut)

Solve for x:... (answered by edjones)

solve logbase2(x+3)+... (answered by lwsshak3)

Please help me solve for x: a) logbase12(x^2-x)=1 b) (logbase2 x)^2-6*(logbase2 x)... (answered by edjones)

Solve for x: logbase2(logbase3(logbase4... (answered by stanbon,josmiceli)