which equation has x = 4 as the solution?\nlog₄(3x + 4)=2\nlog₃(2x - 5)=2\nlogₓ64 = 4\nlogₓ16 = 4

Explanation:

Step1: Recall log - exponent conversion

If $\log_{a}b = c$, then $a^{c}=b$.

Step2: Check $\log_{4}(3x + 4)=2$

Using the conversion, we have $4^{2}=3x + 4$. So, $16=3x + 4$. Subtracting 4 from both sides gives $12 = 3x$, and $x = 4$.

Step3: Check $\log_{3}(2x - 5)=2$

Converting, $3^{2}=2x - 5$, so $9=2x - 5$. Adding 5 gives $14 = 2x$, and $x = 7$.

Step4: Check $\log_{x}64 = 4$

Converting, $x^{4}=64$. Taking the fourth - root of both sides, $x=\sqrt[4]{64}\neq4$.

Step5: Check $\log_{x}16 = 4$

Converting, $x^{4}=16$. Taking the fourth - root of both sides, $x = 2$.

Answer:

$\log_{4}(3x + 4)=2$