Question 334146
{{{log(4, (3x+1)) = 2}}}
When the equation is logarithmic and the variable you are solving for is in the argument of the logarithm, the key is to rewrite the logarithmic equation in exponential form:
{{{4^2 = 3x+1}}}
which simplifies to:
{{{16 = 3x + 1}}}
This is now a very simple equation to solve:
{{{15 = 3x}}}
{{{5 = x}}}
With any logarithmic equation you should check your answers. You must ensure that no arguments of any logarithms become zero or negative!<br>
Checking x = 5 in the original equation:
{{{log(4, (3x+1)) = 2}}}
{{{log(4, (3(5)+1)) = 2}}}
{{{log(4, (15+1)) = 2}}}
{{{log(4, (16)) = 2}}}
The argument is positive to the solution checks!