Simplify the Product: 1/(ln 4) × 1/(log₈ 10) Logarithmic Expression

Change of Base Formula with Complex Products

1ln41log810= \frac{1}{\ln4}\cdot\frac{1}{\log_810}=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:04 Convert from radian to log
00:13 Make sure to multiply numerator by numerator and denominator by denominator
00:20 We'll use the formula for log multiplication, switch between bases
00:36 We'll use the formula for 1 divided by log, we'll get the inverse log
00:41 We'll use this formula in our exercise
00:56 We'll solve the log, and substitute in the exercise
01:16 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

1ln41log810= \frac{1}{\ln4}\cdot\frac{1}{\log_810}=

2

Step-by-step solution

To solve the problem, we must evaluate the expression 1ln41log810\frac{1}{\ln 4} \cdot \frac{1}{\log_8 10}.

First, convert log810\log_8 10 using the change of base formula. We have:

  • log810=ln10ln8\log_8 10 = \frac{\ln 10}{\ln 8}.

Substitute this back into the original expression:

1ln41log810=1ln4ln8ln10\frac{1}{\ln 4} \cdot \frac{1}{\log_8 10} = \frac{1}{\ln 4} \cdot \frac{\ln 8}{\ln 10}.

Next, we need to simplify the expression. We know that ln8=ln(23)=3ln2\ln 8 = \ln (2^3) = 3 \ln 2 and ln4=ln(22)=2ln2\ln 4 = \ln (2^2) = 2 \ln 2.

Substitute these into the expression:

= 12ln23ln2ln10\frac{1}{2 \ln 2} \cdot \frac{3 \ln 2}{\ln 10}.

Simplify by canceling ln2\ln 2:

= 321ln10\frac{3}{2} \cdot \frac{1}{\ln 10}.

Now express ln10=ln(eloge)\ln 10 = \ln (e \cdot \log e), meaning this is equivalent to loge\log e. Continuing, the expression 321loge=32loge\frac{3}{2} \cdot \frac{1}{\log e} = \frac{3}{2} \log e.

Therefore, the simplified solution to the given expression is 32loge\frac{3}{2} \log e.

3

Final Answer

32loge \frac{3}{2}\log e

Key Points to Remember

Essential concepts to master this topic
  • Change of Base: Convert log810=ln10ln8 \log_8 10 = \frac{\ln 10}{\ln 8} to use natural logarithms
  • Power Rules: Simplify ln8=3ln2 \ln 8 = 3\ln 2 and ln4=2ln2 \ln 4 = 2\ln 2 to cancel terms
  • Check: Verify 32loge=32ln10 \frac{3}{2}\log e = \frac{3}{2 \ln 10} equals original expression ✓

Common Mistakes

Avoid these frequent errors
  • Incorrectly applying logarithm properties to products
    Don't treat 1ln41log810 \frac{1}{\ln 4} \cdot \frac{1}{\log_8 10} as ln(14log810) \ln(\frac{1}{4 \cdot \log_8 10}) ! This misapplies logarithm rules to multiplication of fractions, not logarithms of products. Always use change of base formula first, then simplify the algebraic expression.

Practice Quiz

Test your knowledge with interactive questions

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

FAQ

Everything you need to know about this question

Why do I need to use the change of base formula here?

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The change of base formula lets you convert different logarithm bases to the same base. Since we have ln4 \ln 4 (base e) and log810 \log_8 10 (base 8), we need a common base to combine them effectively.

How do I know when to use power rules for logarithms?

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Use power rules when you see logarithms of numbers that are perfect powers! Since 8=23 8 = 2^3 and 4=22 4 = 2^2 , you can write ln8=3ln2 \ln 8 = 3\ln 2 and ln4=2ln2 \ln 4 = 2\ln 2 to simplify.

What does loge \log e mean in the final answer?

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loge \log e represents the common logarithm of e (base 10). Since 1ln10=loge \frac{1}{\ln 10} = \log e , this is just another way to express the reciprocal of the natural logarithm of 10.

Can I solve this problem using a different base?

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Yes! You could convert everything to base 10 or base 2 instead of natural logarithms. The key is being consistent and using the same base throughout your solution.

Why does ln2 \ln 2 cancel out in the middle steps?

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When you have 12ln23ln2ln10 \frac{1}{2\ln 2} \cdot \frac{3\ln 2}{\ln 10} , the ln2 \ln 2 terms are in the numerator and denominator, so they divide out: 3ln22ln2=32 \frac{3\ln 2}{2\ln 2} = \frac{3}{2} .

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