Solve the Logarithmic Inequality: Determine x in log_4(7) + log_4(2) ≤ log_4(x)

Question

$\log_47+\log_42\le\log_4x$

$x=\text{?}$

To solve the given inequality $\log_4(7) + \log_4(2) \le \log_4(x)$ , we will utilize the properties of logarithms:

First, apply the logarithm sum property: $\log_4(7) + \log_4(2) = \log_4(7 \times 2) = \log_4(14)$ .
Now, the inequality becomes $\log_4(14) \le \log_4(x)$ .
Since the logarithm function is monotonically increasing when the base is greater than 1, we can simplify the inequality to $14 \le x$ .

Therefore, the solution to the inequality is $x \ge 14$ .

Therefore, the correct choice is $14 \le x$ , which matches the given correct answer.

$14\le x$