Solve 2log₃8: Step-by-Step Logarithm Equation Solution

Logarithm Properties with Power Rules

2log38= 2\log_38=

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Step-by-step video solution

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00:00 Solve
00:03 We'll use the formula for the logarithm of a power
00:11 We'll use this formula in our exercise
00:21 Let's calculate the power
00:25 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

2log38= 2\log_38=

2

Step-by-step solution

To solve this problem, let's simplify 2log382\log_3 8 using logarithm rules.

  • Step 1: Recognize the expression form
    The expression is of the form alogbca \cdot \log_b c, where a=2a = 2, b=3b = 3, and c=8c = 8.
  • Step 2: Apply the power property
    According to the power property of logarithms, 2log382 \cdot \log_3 8 can be simplified to log3(82)\log_3 (8^2).
  • Perform the calculation
    Calculate 828^2, which is 6464.
  • Step 3: Simplify further
    Therefore, we have log364\log_3 64.

This is a straightforward application of the power property of logarithms. By applying this property correctly, we've simplified the original expression correctly.

Therefore, the simplified form of 2log382\log_3 8 is log364\log_3 64.

3

Final Answer

log364 \log_364

Key Points to Remember

Essential concepts to master this topic
  • Power Property: Coefficient becomes exponent inside logarithm argument
  • Technique: Transform 2log38 2\log_3 8 to log3(82) \log_3(8^2) then calculate 82=64 8^2 = 64
  • Check: Verify 2log38=log364 2\log_3 8 = \log_3 64 using calculator approximations ✓

Common Mistakes

Avoid these frequent errors
  • Moving coefficient to the base instead of argument
    Don't change 2log38 2\log_3 8 to log68 \log_6 8 = wrong base! The coefficient multiplies the entire logarithm, not the base. Always move the coefficient as an exponent to the argument inside the log.

Practice Quiz

Test your knowledge with interactive questions

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

FAQ

Everything you need to know about this question

Why does the coefficient 2 become an exponent?

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This comes from the power property of logarithms: alogbc=logb(ca) a \cdot \log_b c = \log_b(c^a) . The coefficient always moves inside as an exponent on the argument, not the base!

What if I have a fraction as the coefficient?

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Same rule applies! For example, 12log38=log3(81/2)=log3(8) \frac{1}{2}\log_3 8 = \log_3(8^{1/2}) = \log_3(\sqrt{8}) . The fraction becomes the exponent inside.

Can I work backwards from the answer choices?

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Yes! Since log364 \log_3 64 means "3 to what power equals 64?", you can verify this equals 2log38 2\log_3 8 using the power property.

How do I remember which direction the power property goes?

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Think: "Coefficient goes inside as an exponent." The number outside the log always moves to become an exponent on the argument inside the parentheses.

What if the base changes in the answer choices?

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Be careful! The base never changes when using the power property. If you see log68 \log_6 8 as an option, it's likely a trap answer.

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