Question 816636
{{{log(5, (3x+1))- log(5,( 4))=2}}}
First, use the {{{log(a,(p))-log(a, (q)) = log(a, (p/q))}}} property to combine the logs:
{{{log(5, ((3x+1)/4))=2}}}
Then rewrite the equation in exponential form. In general {{{log(a, (p)) = n}}} is equivalent to {{{p = a^n}}}. Using this pattern on our equation we get:
{{{(3x+1)/4 = 5^2}}}
which simplifies to:
{{{(3x+1)/4 = 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<br>
Last, we check. This is <i>not</i> 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.<br>
Use the original equation to check:
{{{log(5, (3x+1))- log(5,( 4))=2}}}
Checking x = 33:
{{{log(5, (3(33)+1))- log(5,( 4))=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!