SOLUTION: 1. use laws of logarithm to express each side of the equation as a single logarithm. Then compare both sides of the equation to solve: log4(4's the base)X + log4(4's the base)12= l

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: 1. use laws of logarithm to express each side of the equation as a single logarithm. Then compare both sides of the equation to solve: log4(4's the base)X + log4(4's the base)12= l      Log On


   

Question 826267: 1. use laws of logarithm to express each side of the equation as a single logarithm. Then compare both sides of the equation to solve: log4(4's the base)X + log4(4's the base)12= log4(4 is base)48
2. 1+log3(3 is the base)x^2

Thanks in advance!
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
log%284%2C+%28x%29%29+%2B+log%284%2C+%2812%29%29+=+log%284%2C+%2848%29%29
First we use a property of logarithms, log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29, to combine the two logs on the left:
log%284%2C+%28x%2A12%29%29+=+log%284%2C+%2848%29%29
or
log%284%2C+%2812x%29%29+=+log%284%2C+%2848%29%29
The equation now says that two base 4 logs are equal. The only way this can be true is if the arguments of both logs are equal, too. So:
4x = 12
Dividing by 4:
x = 3

Checking is not optional when solving logarithmic equations like this. We must check to ensure that all bases and arguments are valid for the solution(s). (Valid bases are positive but not 1 and valid arguments are positive. Use the original equation to check:
log%284%2C+%28x%29%29+%2B+log%284%2C+%2812%29%29+=+log%284%2C+%2848%29%29
Checking x = 3:
log%284%2C+%283%29%29+%2B+log%284%2C+%2812%29%29+=+log%284%2C+%2848%29%29
We can already see that all bases are 4 (valid) and all the arguments are positive (valid). So x = 3 checks out and is the solution.

P.S. If x = 3 had made any base or any argument of any logarithm invalid, we would have to reject it, even if it is the only "solution" we found.