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

log47+log42log4x \log_47+\log_42\le\log_4x

x=? x=\text{?}

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1

Understand the problem

log47+log42log4x \log_47+\log_42\le\log_4x

x=? x=\text{?}

2

Step-by-step solution

To solve the given inequality log4(7)+log4(2)log4(x) \log_4(7) + \log_4(2) \le \log_4(x) , we will utilize the properties of logarithms:

  • First, apply the logarithm sum property: log4(7)+log4(2)=log4(7×2)=log4(14) \log_4(7) + \log_4(2) = \log_4(7 \times 2) = \log_4(14) .
  • Now, the inequality becomes log4(14)log4(x) \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 14x 14 \le x .

Therefore, the solution to the inequality is x14 x \ge 14 .

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

3

Final Answer

14x 14\le x

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\( \frac{1}{\log_49}= \)

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