help solve for x

Question

1. log base3(x+4)+log base3(x-1)=3
2. log base2(2x^2-x+3)=2
3. 2log base2(2x-1)-2log base2(x+1)=2

Answer 1

1. log[3](x+4)+log[3](x-1)=log[3](x^2+3x-4)=3. (x^2+3x-4)=3^3=27; x^2+3x-31=0; x=(-3±√133)/2, x=4.2663 approx only, because the negative root would give us negative values for the log arguments, which are disallowed.
2. 2x^2-x+3=4, raising each side as a power of 2; 2x^2-x-1=0; (2x+1)(x-1)=0, so x=1 or -1/2.
3. Divide through by 2 and combine the logs: log[2]((2x-1)/(x+1))=1; (2x-1)/(x+1)=2; 2x-1=2x+2 has no solutions, because -1≠2.