# SOLUTION: log(6x+3)=1+log(x-6)

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 Question 605412: log(6x+3)=1+log(x-6)Found 2 solutions by stanbon, bucky:Answer by stanbon(57377)   (Show Source): You can put this solution on YOUR website!log(6x+3)=1+log(x-6) ---- log(6x+3) - log(x-6) = 1 ---- log[(6x+3)/(x-6)] = 1 ---- (6x+3)/(x-6) = 10 --- 6x+3 = 10x-60 4x = 63 x = 63/4 ============== cheers, Stan H. ================= Answer by bucky(2189)   (Show Source): You can put this solution on YOUR website!Given to solve for x . . Subtract log(x-6) from both sides to get: . . By the rules of logarithms, the subtraction of two logarithms is the same as the logarithm of their division with the divisor being the quantity in the negative logarithm. So we can writ that the above equation is equivalent to: . . Next convert this to exponential form by raising the base (base 10) of the logarithm to the power which is on the right side of the equal sign and setting this equal to the quantity that the logarithm operator is acting on. This results in: . . The 10 to the first power is just 10. So the equation is: . . Get rid of the denominator on the left side by multiplying both sides by x - 6: . . . Do the distributed multiplication on the right side by multiplying 10 times each of the two terms in the parentheses, and the result is: . . Subtract 10x from both sides: . . Subtract 3 from both sides: . . Multiply both sides by -1: . . Solve for x by dividing both sides by 4 and you arrive at the solution: . . . Hope this helps you to understand how this problem can be worked by using the properties of logarithms in a logical way. .