Solve log₄x > 7log₄2: Base-4 Logarithmic Inequality

Logarithmic Inequalities with Power Properties

7log42<log4x 7\log_42<\log_4x

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Step-by-step video solution

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00:00 Solve
00:05 We will use the logarithm formula of power, raise the number by the coefficient
00:25 We will equate the logarithms
00:34 And this is the solution to the question

Step-by-step written solution

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1

Understand the problem

7log42<log4x 7\log_42<\log_4x

2

Step-by-step solution

To solve the inequality 7log42<log4x7\log_4 2 < \log_4 x, we will follow these steps:

  • Step 1: Simplify 7log427\log_4 2 using the logarithm properties.
  • Step 2: Write the inequality log4(27)<log4x\log_4(2^7) < \log_4 x.
  • Step 3: Solve for xx by converting the logarithmic inequality into an exponential form.

Let's now proceed with these steps:

Step 1: Using the power property of logarithms, we have 7log42=log4(27)7\log_4 2 = \log_4 (2^7). This step simplifies the multiplication into a single logarithmic term.

Step 2: Using substitution in the inequality, we write it as log4(27)<log4x\log_4 (2^7) < \log_4 x.

Step 3: Since logarithms are one-to-one functions, we can conclude that if log4(27)<log4x\log_4 (2^7) < \log_4 x, then 27<x2^7 < x. This results from the property ac=ad    c=da^c = a^d \iff c = d where the bases are equal.

Therefore, the solution to the inequality is 27<x\mathbf{2^7 < x}.

3

Final Answer

27<x 2^7 < x

Key Points to Remember

Essential concepts to master this topic
  • Power Rule: Convert 7log₄2 to log₄(2⁷) using logarithm properties
  • Technique: Use one-to-one property: if log₄A < log₄B, then A < B
  • Check: Verify 2⁷ = 128, so x > 128 makes log₄x > log₄128 ✓

Common Mistakes

Avoid these frequent errors
  • Incorrectly applying exponent to the base instead of the argument
    Don't write 7log₄2 as log₇₄2 or think it equals 7⁴! This confuses where the exponent belongs and gives completely wrong results. Always apply the coefficient as an exponent to the argument inside the logarithm: 7log₄2 = log₄(2⁷).

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 remove the logarithms from both sides of the inequality?

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Since logarithmic functions are one-to-one and strictly increasing, they preserve the inequality direction. If log4A<log4B \log_4 A < \log_4 B , then we know A<B A < B .

How do I know which power property to use?

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When you see a coefficient in front of a logarithm like 7log42 7\log_4 2 , use the power rule: nlogba=logb(an) n\log_b a = \log_b(a^n) . The coefficient becomes an exponent!

What if the base of the logarithm was different on each side?

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If the bases are different, you cannot directly compare the arguments. You would need to convert to the same base using change of base formula first.

Do I need to calculate 2⁷ to solve this?

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Not necessarily! The answer x>27 x > 2^7 is perfectly correct. However, calculating 27=128 2^7 = 128 can help you verify your work and understand the solution better.

Why doesn't the inequality direction change?

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The inequality direction only flips when you multiply or divide by a negative number. Since we're using logarithm properties and the one-to-one property, no negative multiplication occurs!

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