Simplify the Expression: log₉(e²)/log₉(e) Using Log Properties

Q: Why can't I use the quotient rule of logarithms here?

The quotient rule applies to $ \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) $. But here we have a fraction of logarithms, not a logarithm of a fraction! Use algebraic division instead.

Q: Does this work with any exponent, not just 2?

Yes! The pattern is $ \frac{\log_b(a^n)}{\log_b(a)} = \frac{n \cdot \log_b(a)}{\log_b(a)} = n $. The answer is always the exponent, regardless of the base or the number inside the log.

Q: Why does log₉(e) cancel out completely?

Think of it like regular fractions! When you have $ \frac{2x}{x} $, the x's cancel to give 2. Here, $ \frac{2 \cdot \log_9(e)}{\log_9(e)} $ works the same way - the $ \log_9(e) $ terms cancel.

Q: What if I don't remember the power rule?

Remember: exponents become multipliers in logarithms. $ \log(a^3) = 3 \cdot \log(a) $, $ \log(x^5) = 5 \cdot \log(x) $. The exponent always comes down in front!

Logarithmic Division with Power Rules

$\frac{\log_9e^2}{\log_9e}=$

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Step-by-step video solution

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00:00 Solve

00:03 We'll use the formula for logarithmic division

00:08 We'll get the logarithm of the numerator with the denominator as the base

00:13 We'll use this formula in our exercise

00:23 We'll use the formula for logarithm of a power

00:28 We'll use this formula in our exercise

00:40 The logarithm of any number in its own base is always equal to 1

00:53 And this is the solution to the question

Step-by-step written solution

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Understand the problem

$\frac{\log_9e^2}{\log_9e}=$

Step-by-step solution

To solve this problem, we'll simplify the given expression $\frac{\log_9e^2}{\log_9e}$ using logarithmic rules:

Step 1: Apply the power rule of logarithms:
The numerator $\log_9e^2$ can be rewritten using the power rule: $\log_9e^2 = 2 \cdot \log_9e$ .

Step 2: Substitute and simplify the fraction:
Substitute the expression from Step 1 into the original problem:
$\frac{\log_9e^2}{\log_9e} = \frac{2 \cdot \log_9e}{\log_9e}$ .

Step 3: Cancel common terms:
Since $\log_9e$ appears in both the numerator and the denominator, it cancels out, leaving:
$\frac{2 \cdot \log_9e}{\log_9e} = 2$ .

Therefore, the solution to the problem is $\boxed{2}$ .

Final Answer

$2$

Key Points to Remember

Essential concepts to master this topic

Power Rule: log_b(a^n) = n · log_b(a) brings exponents down
Technique: Convert log₉(e²) to 2·log₉(e) using power rule
Check: Substitute back: 2·log₉(e)/log₉(e) = 2·1 = 2 ✓

Common Mistakes

Avoid these frequent errors

Trying to subtract exponents instead of using power rule
Don't think log₉(e²)/log₉(e) = log₉(e²⁻¹) = log₉(e) = some complex answer! This confuses quotient rules with simple fraction division. Always apply the power rule first: log₉(e²) = 2·log₉(e), then divide.

Practice Quiz

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$\log_{10}3+\log_{10}4=$

FAQ

Everything you need to know about this question

Why can't I use the quotient rule of logarithms here?

The quotient rule applies to $\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)$ . But here we have a fraction of logarithms, not a logarithm of a fraction! Use algebraic division instead.

What if the bases were different in numerator and denominator?

If the bases don't match, you can't simplify this way! You'd need to convert to the same base using change of base formula first, or use properties like $\log_a(b) = \frac{1}{\log_b(a)}$ .

Does this work with any exponent, not just 2?

Yes! The pattern is $\frac{\log_b(a^n)}{\log_b(a)} = \frac{n \cdot \log_b(a)}{\log_b(a)} = n$ . The answer is always the exponent, regardless of the base or the number inside the log.

Why does log₉(e) cancel out completely?

Think of it like regular fractions! When you have $\frac{2x}{x}$ , the x's cancel to give 2. Here, $\frac{2 \cdot \log_9(e)}{\log_9(e)}$ works the same way - the $\log_9(e)$ terms cancel.

What if I don't remember the power rule?

Remember: exponents become multipliers in logarithms. $\log(a^3) = 3 \cdot \log(a)$ , $\log(x^5) = 5 \cdot \log(x)$ . The exponent always comes down in front!

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Simplify the Expression: log₉(e²)/log₉(e) Using Log Properties

Logarithmic Division with Power Rules

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why can't I use the quotient rule of logarithms here?

What if the bases were different in numerator and denominator?

Does this work with any exponent, not just 2?

Why does log₉(e) cancel out completely?

What if I don't remember the power rule?

🌟 Unlock Your Math Potential

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