Finding the Largest Value: Comparing Numbers with Constraint a>1

Q: Why does a>1 matter for comparing these expressions?

When a > 1, larger exponents give larger values. For example, if a = 2, then $ 2^3 = 8 $ while $ 2^2 = 4 $. If a

Q: How do I handle negative exponents in comparisons?

Negative exponents create fractions: $ a^{-12} = \frac{1}{a^{12}} $. Since a > 1, this becomes a very small positive fraction, much smaller than any positive power of a.

Q: What if I forget the exponent division rule?

Remember: $ \frac{a^m}{a^n} = a^{m-n} $. Think of it as canceling common factors. You can also write it as $ a^m \times a^{-n} = a^{m-n} $ using multiplication rules.

Q: Can I use specific numbers to check my answer?

Absolutely! Try a = 2: $ \frac{2^{12}}{2^{10}} = 2^2 = 4 $, $ \frac{2^4}{2^2} = 2^2 = 4 $, $ \frac{2^{-7}}{2^5} = 2^{-12} \approx 0.0002 $, and $ 2^3 = 8 $. Clearly 8 is largest!

Q: Why are the first two options both equal to a²?

This happens because $ 12 - 10 = 2 $ and $ 4 - 2 = 2 $. Different-looking fractions can simplify to the same result! Always simplify completely before comparing.

Exponent Laws with Positive Base Constraints

Which value is the largest?

given that $a>1$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Identify the largest value

00:03 When dividing powers with equal bases

00:07 The power of the result equals the difference between the powers

00:12 We'll apply this formula to our exercise and subtract the powers

00:18 We'll use this method to solve all sections

00:44 Identify the largest power

00:53 This is the solution

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Which value is the largest?

given that $a>1$

Step-by-step solution

Note that in almost all options there are fractions where both the numerator and denominator have identical bases, therefore we will use the division law between terms with identical bases to solve the exercise:

$\frac{c^m}{c^n}=c^{m-n}$ Let's apply this to the problem, first we'll simplify each of the given options using the above law (options in order):

$\frac{a^{12}}{a^{10}}=a^{12-10}=a^2$ $\frac{a^4}{a^2}=a^{4-2}=a^2$ $\frac{a^{-7}}{a^5}=a^{-7-5}=a^{-12}$ $a^3$ Back to the problem, given that:

$a>1$ therefore the option with the largest value will be the one where $a$ has the largest exponent (for emphasis - a positive exponent is greater than a negative exponent),

meaning the option:

$a^3$ above is correct,

therefore answer D is correct.

Final Answer

$a^3$

Key Points to Remember

Essential concepts to master this topic

Division Rule: When dividing powers with same base, subtract exponents
Simplification: $\frac{a^{12}}{a^{10}} = a^{12-10} = a^2$ , $\frac{a^4}{a^2} = a^2$
Comparison: Since a>1, higher positive exponent gives largest value: $a^3 > a^2 > a^{-12}$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting to apply exponent division rules before comparing
Don't compare fractions like $\frac{a^{12}}{a^{10}}$ directly without simplifying = wrong comparison! The fractions look complex but simplify to basic powers. Always apply $\frac{a^m}{a^n} = a^{m-n}$ first, then compare exponents.

Practice Quiz

Test your knowledge with interactive questions

$(3\times4\times5)^4=$

FAQ

Everything you need to know about this question

Why does a>1 matter for comparing these expressions?

When a > 1, larger exponents give larger values. For example, if a = 2, then $2^3 = 8$ while $2^2 = 4$ . If a < 1, this relationship would flip!

How do I handle negative exponents in comparisons?

Negative exponents create fractions: $a^{-12} = \frac{1}{a^{12}}$ . Since a > 1, this becomes a very small positive fraction, much smaller than any positive power of a.

What if I forget the exponent division rule?

Remember: $\frac{a^m}{a^n} = a^{m-n}$ . Think of it as canceling common factors. You can also write it as $a^m \times a^{-n} = a^{m-n}$ using multiplication rules.

Can I use specific numbers to check my answer?

Absolutely! Try a = 2: $\frac{2^{12}}{2^{10}} = 2^2 = 4$ , $\frac{2^4}{2^2} = 2^2 = 4$ , $\frac{2^{-7}}{2^5} = 2^{-12} \approx 0.0002$ , and $2^3 = 8$ . Clearly 8 is largest!

Why are the first two options both equal to a²?

This happens because $12 - 10 = 2$ and $4 - 2 = 2$ . Different-looking fractions can simplify to the same result! Always simplify completely before comparing.

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