ANSWER
[tex] \log_{4}(3x + 4) = 2[/tex]
EXPLANATION
Consider the equation:
[tex] \log_{4}(3x + 4) = 2[/tex]
When we rewrite this logarithmic equation in the exponential form, we obtain:
[tex]3x + 4= {4}^{2} [/tex]
Note that to write a logarithmic equation in exponential form, the base of the logarithm is still the base in the exponential form.
We now simplify the RHS.
[tex]3x + 4 = 16[/tex]
Group like terms
[tex]3x = 16 - 4[/tex]
This implies that
[tex]3x = 12[/tex]
Divide both sides by 3
[tex] \frac{3x}{3} = \frac{12}{3} [/tex]
Simplify to get;
[tex]x = 4[/tex]
Hence the equation that has x=4 as a solution is
[tex] \log_{4}(3x + 4) = 2[/tex]
Another way to do this is to substitute x=4 into each equation. The equation that is satisfied is the correct choice.
