SOLUTION: Can you help me solve this problem Solve. Log5 (3x+1)- Log5 4=2

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Question 816636: Can you help me solve this problem
Solve. Log5 (3x+1)- Log5 4=2
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%285%2C+%283x%2B1%29%29-+log%285%2C%28+4%29%29=2
First, use the log%28a%2C%28p%29%29-log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29 property to combine the logs:
log%285%2C+%28%283x%2B1%29%2F4%29%29=2
Then rewrite the equation in exponential form. In general log%28a%2C+%28p%29%29+=+n is equivalent to p+=+a%5En. Using this pattern on our equation we get:
%283x%2B1%29%2F4+=+5%5E2
which simplifies to:
%283x%2B1%29%2F4+=+25
Now we solve this. Multiplying each side by 4 (to eliminate the fraction):
3x + 1 = 100
Subtracting 1:
3x = 99
Dividing by 3:
x = 33

Last, we check. This is not optional! A check must be made to ensure that the "solution(s)" make all bases and arguments of all logs valid. (Valid bases are any positive number except 1 and valid arguments are positive.) Any "solution" that makes any base or any argument invalid must be rejected.

Use the original equation to check:
log%285%2C+%283x%2B1%29%29-+log%285%2C%28+4%29%29=2
Checking x = 33:
log%285%2C+%283%2833%29%2B1%29%29-+log%285%2C%28+4%29%29=2
We can already see that the bases are 5's and the arguments are (or will end up) positive -- both valid. So this solution checks!