Solve: 2log₃4 × log₂9 Logarithm Product Problem

Logarithm Products with Change of Base

2log34×log29= 2\log_34\times\log_29=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:07 Let's solve this problem together.
00:12 We'll start by using the formula for the logarithm of a power. Ready?
00:20 First, let's calculate the power. Here we go!
00:24 Now, we'll apply the formula for multiplying logarithms.
00:30 Next, we'll change the bases of the logarithms. Just follow along.
00:38 We'll use this formula in our exercise. Let's continue.
00:57 Let's calculate the first logarithm now.
01:10 Great! Now, let's do the same for the second logarithm.
01:19 Let's substitute our solutions and finish the calculation.
01:24 And this is how we solve the problem. Well done!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

2log34×log29= 2\log_34\times\log_29=

2

Step-by-step solution

To solve this problem, we need to evaluate 2log34×log29 2\log_3 4 \times \log_2 9 . We'll use the change of base formula to simplify the logarithms.

  • Step 1: Apply the change of base formula to both logarithms.
  • Step 2: Simplify the expressions by substituting appropriate values.
  • Step 3: Compute the multiplication of the simplified values.

Step 1: Convert the logarithms using the change of base formula:

log34=log104log103\log_3 4 = \frac{\log_{10} 4}{\log_{10} 3} and log29=log109log102\log_2 9 = \frac{\log_{10} 9}{\log_{10} 2}.

Step 2: Substitute these back into the expression:

2×log104log103×log109log1022 \times \frac{\log_{10} 4}{\log_{10} 3} \times \frac{\log_{10} 9}{\log_{10} 2}.

Recognize that log104=2log102\log_{10} 4 = 2 \log_{10} 2 and log109=2log103\log_{10} 9 = 2 \log_{10} 3, hence simplifying gives:

= 2×2log102log103×2log103log1022 \times \frac{2 \log_{10} 2}{\log_{10} 3} \times \frac{2 \log_{10} 3}{\log_{10} 2}.

Step 3: Cancel terms and calculate:

The terms log102\log_{10} 2 and log103\log_{10} 3 cancel out:

= 2×2×2=82 \times 2 \times 2 = 8.

Therefore, the solution to the problem is 8 \boxed{8} , which corresponds to choice 3 in the provided answer choices.

3

Final Answer

8 8

Key Points to Remember

Essential concepts to master this topic
  • Change of Base: Use logab=logclogc \log_a b = \frac{\log c}{\log c} to convert bases
  • Power Property: Recognize log4=2log2 \log 4 = 2\log 2 and log9=2log3 \log 9 = 2\log 3
  • Verification: Cancel matching terms and multiply coefficients: 2×2×2 = 8 ✓

Common Mistakes

Avoid these frequent errors
  • Not using change of base formula
    Don't try to evaluate log34 \log_3 4 and log29 \log_2 9 directly = getting stuck with irrational decimals! This makes calculation nearly impossible without a calculator. Always convert to common base using change of base formula first.

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't I just use a calculator for the logarithms?

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While calculators work, understanding the change of base method helps you see the beautiful cancellation pattern! Plus, many tests require exact answers, not decimal approximations.

Which base should I convert to?

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Any base works! Base 10 or natural log (ln) are most common. The key is using the same base for both logarithms so terms can cancel.

How do I remember that log 4 = 2 log 2?

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Remember the power property: log(an)=nloga \log(a^n) = n\log a . Since 4=22 4 = 2^2 , we get log4=log(22)=2log2 \log 4 = \log(2^2) = 2\log 2 !

What if the logarithms don't cancel perfectly?

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That's normal for many problems! The beauty of this specific problem is the perfect cancellation. In general cases, you'd simplify as much as possible using logarithm properties.

Can I solve this without change of base?

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Technically yes, but it's much harder! You'd need to use properties like logablogbc=logac \log_a b \cdot \log_b c = \log_a c , but change of base is the clearest method for this type of problem.

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