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

Logarithmic Inequalities with Product Property

log⁡47+log⁡42≤log⁡4x \log_47+\log_42\le\log_4x

x=? x=\text{?}

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1

Understand the problem

log⁡47+log⁡42≤log⁡4x \log_47+\log_42\le\log_4x

x=? x=\text{?}

2

Step-by-step solution

To solve the given inequality log⁡4(7)+log⁡4(2)≤log⁡4(x) \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(7×2)=log⁡4(14) \log_4(7) + \log_4(2) = \log_4(7 \times 2) = \log_4(14) .
  • Now, the inequality becomes log⁡4(14)≤log⁡4(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 14≤x 14 \le x .

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

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

3

Final Answer

14≤x 14\le x

Key Points to Remember

Essential concepts to master this topic
  • Product Property: log⁡a(m)+log⁡a(n)=log⁡a(mn) \log_a(m) + \log_a(n) = \log_a(mn)
  • Technique: log⁡4(7)+log⁡4(2)=log⁡4(14) \log_4(7) + \log_4(2) = \log_4(14) using multiplication inside
  • Check: Since base 4 > 1, log⁡4(14)≤log⁡4(x) \log_4(14) \le \log_4(x) means 14≤x 14 \le x ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting domain restrictions for logarithms
    Don't just solve 14≤x 14 \le x and forget x > 0! This gives incomplete solutions like negative values. The logarithm log⁡4(x) \log_4(x) requires x to be positive. Always state the complete solution: x≥14 x \ge 14 (which automatically satisfies x > 0).

Practice Quiz

Test your knowledge with interactive questions

\( \log_{10}3+\log_{10}4= \)

FAQ

Everything you need to know about this question

Why can I combine the logarithms on the left side?

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The product property says log⁡a(m)+log⁡a(n)=log⁡a(mn) \log_a(m) + \log_a(n) = \log_a(mn) . So log⁡4(7)+log⁡4(2)=log⁡4(7×2)=log⁡4(14) \log_4(7) + \log_4(2) = \log_4(7 \times 2) = \log_4(14) . This only works when the bases are the same!

How do I remove the logarithms from both sides?

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Since the logarithm function with base > 1 is increasing, if log⁡4(14)≤log⁡4(x) \log_4(14) \le \log_4(x) , then 14≤x 14 \le x . The inequality direction stays the same!

Do I need to worry about x being positive?

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Yes, but since our answer is x≥14 x \ge 14 , and 14 > 0, we automatically satisfy the requirement that x must be positive for log⁡4(x) \log_4(x) to exist.

What if the base was between 0 and 1?

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If the base were between 0 and 1, the logarithm function would be decreasing, so you'd need to flip the inequality sign when removing the logarithms. Always check if your base is greater than or less than 1!

Can I check my answer by substituting?

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Absolutely! Try x=14 x = 14 : log⁡4(7)+log⁡4(2)=log⁡4(14)≤log⁡4(14) \log_4(7) + \log_4(2) = \log_4(14) \le \log_4(14) ✓. Try x=20 x = 20 : log⁡4(14)≤log⁡4(20) \log_4(14) \le \log_4(20) ✓ since 14 < 20.

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