Solve the Logarithm Ratio: Finding log₈(9a)/log₈(3a)

Logarithm Quotient Rule with Base Change

log89alog83a= \frac{\log_89a}{\log_83a}=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:03 We will use the formula for logical division
00:08 We will get the log of the numerator with the denominator as the base
00:16 We will use this formula in our exercise
00:21 And this is the solution to the question

Step-by-step written solution

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1

Understand the problem

log89alog83a= \frac{\log_89a}{\log_83a}=

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Apply the mathematical property using the quotient rule for logarithms.
  • Step 2: Simplify the expression into the desired format.

Now, let's work through each step:
Step 1: Given the expression log89alog83a\frac{\log_8 9a}{\log_8 3a}, we can directly apply the quotient rule for logarithms, which tells us that logbMlogbN=logNM\frac{\log_b M}{\log_b N} = \log_N M.
Step 2: Applying this formula, we find that log89alog83a=log3a9a\frac{\log_8 9a}{\log_8 3a} = \log_{3a} 9a.

Therefore, the solution to the problem is log3a9a \log_{3a} 9a .

3

Final Answer

log3a9a \log_{3a}9a

Key Points to Remember

Essential concepts to master this topic
  • Quotient Rule: logbMlogbN=logNM \frac{\log_b M}{\log_b N} = \log_N M transforms division into base change
  • Technique: Apply formula directly: log89alog83a=log3a9a \frac{\log_8 9a}{\log_8 3a} = \log_{3a} 9a
  • Check: Verify the base becomes denominator and argument becomes numerator ✓

Common Mistakes

Avoid these frequent errors
  • Trying to simplify using regular logarithm properties
    Don't use log8(9a)log8(3a)=log8(9a3a) \log_8(9a) - \log_8(3a) = \log_8(\frac{9a}{3a}) ! This is subtraction, not division of logs. The quotient rule for divided logs is different. Always use logbMlogbN=logNM \frac{\log_b M}{\log_b N} = \log_N M when dividing logarithms.

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 cancel out the log₈ parts?

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You cannot cancel logarithms like regular fractions! log89alog83a \frac{\log_8 9a}{\log_8 3a} is not equal to 9a3a \frac{9a}{3a} . Use the special quotient rule for dividing logarithms instead.

How is this different from log properties I know?

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This uses the quotient rule for divided logs, not the quotient property. log(A/B)=logAlogB \log(A/B) = \log A - \log B is for logs of quotients, but logAlogB=logBA \frac{\log A}{\log B} = \log_B A is for quotients of logs!

What if the bases were different?

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The quotient rule only works when both logarithms have the same base. If bases differ, you'd need to use change of base formula first to make them match.

How do I remember which becomes the base and which becomes the argument?

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Remember: denominator becomes base, numerator becomes argument. In log89alog83a \frac{\log_8 9a}{\log_8 3a} , the bottom (3a) becomes the new base, top (9a) stays as argument.

Can I verify this answer somehow?

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Yes! Use the change of base formula: log3a9a=log89alog83a \log_{3a} 9a = \frac{\log_8 9a}{\log_8 3a} . This confirms our answer matches the original expression!

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