# SOLUTION: Ive totally lost myself with this, so if there is anyone that can help please 2log[4](x)+log[4](3)=log[4](x)-log[4](2) All the 4s are log base 4 Thank you

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: Ive totally lost myself with this, so if there is anyone that can help please 2log[4](x)+log[4](3)=log[4](x)-log[4](2) All the 4s are log base 4 Thank you      Log On

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 Click here to see ALL problems on Exponential-and-logarithmic-functions Question 398220: Ive totally lost myself with this, so if there is anyone that can help please 2log[4](x)+log[4](3)=log[4](x)-log[4](2) All the 4s are log base 4 Thank youAnswer by stanbon(57940)   (Show Source): You can put this solution on YOUR website!2log[4](x)+log[4](3)=log[4](x)-log[4](2) ----- log4(x^2) + log4(3) = log4[x/2] --- log4(3x^2) = log4(x/2) --- 3x^2 = x/2 6x^2 = x 6x^2-x = 0 x(6x-1) = 0 --- x cannot be zero, so x = 1/6 is the only answer. --- Let me know if you do not understand all this. Cheers, Stan H. =============