Solve the Exponent Equation: 45^-80 × 1/45^-81 × 49 × 7^-5

Q: Why does $ \frac{1}{45^{-81}} $ become $ 45^{81} $ ?

When you have a negative exponent in the denominator, it flips to become positive in the numerator! Think of it as: $ \frac{1}{a^{-n}} = \frac{1}{\frac{1}{a^n}} = a^n $.

Q: How do I know when to add or subtract exponents?

Multiplying same bases: add exponents ($ a^m \cdot a^n = a^{m+n} $). Dividing same bases: subtract exponents ($ \frac{a^m}{a^n} = a^{m-n} $). In this problem, we're multiplying!

Q: Why does 49 become $ 7^2 $ ?

Because $ 7 \times 7 = 49 $, so $ 49 = 7^2 $. Converting to the same base (base 7) lets us use exponent rules: $ 7^2 \cdot 7^{-5} = 7^{2-5} = 7^{-3} $.

Q: What if I can't recognize that 49 is $ 7^2 $ ?

Practice memorizing perfect squares! Common ones: $ 4=2^2, 9=3^2, 16=4^2, 25=5^2, 36=6^2, 49=7^2, 64=8^2, 81=9^2 $. These patterns appear frequently in exponent problems.

Q: Can I leave my answer as $ \frac{45}{7^3} $ or do I need to calculate $ 7^3 $ ?

Both forms are correct! $ \frac{45}{7^3} $ is the exact form, while $ \frac{45}{343} $ is the decimal form. Check what your teacher prefers - often exact form is better.

Q: How do I check if my final answer is right?

Substitute back into the original expression! Calculate each part: $ 45^{-80} \cdot 45^{81} = 45 $ and $ 49 \cdot 7^{-5} = \frac{1}{7^3} $, then multiply: $ 45 \cdot \frac{1}{7^3} = \frac{45}{7^3} $ ✓

Exponent Rules with Negative Powers

$45^{-80}\cdot\frac{1}{45^{-81}}\cdot49\cdot7^{-5}=\text{?}$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Simplify the following problem

00:03 The number (A) raised to a negative power (N)

00:06 equals 1 divided by the number (A) raised to the same power (N)

00:09 We will apply this formula to our exercise

00:26 Let's break down 49 into 7 squared

00:32 When multiplying powers with equal bases

00:36 The power of the result equals the sum of the powers

00:39 We will apply this formula to our exercise, and then proceed to add together the powers

00:49 Let's calculate the powers

01:01 This is the solution

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

$45^{-80}\cdot\frac{1}{45^{-81}}\cdot49\cdot7^{-5}=\text{?}$

Step-by-step solution

To solve the problem, let's follow these steps:

Step 1: Simplify the expression $45^{-80} \cdot \frac{1}{45^{-81}}$ .
Step 2: Simplify $49 \cdot 7^{-5}$ .
Step 3: Combine results to get the final expression.

Now, let's work through each step:
Step 1: Simplify $45^{-80} \cdot \frac{1}{45^{-81}}$ . Using the exponent rule: $a^{-b} = \frac{1}{a^b}$ and $a^m \cdot a^n = a^{m+n}$ , we have:

$45^{-80} \cdot \frac{1}{45^{-81}} = 45^{-80} \cdot 45^{81} = 45^{-80 + 81} = 45^1 = 45.$

Step 2: Simplify $49 \cdot 7^{-5}$ . Note that $49 = 7^2$ , so we can rewrite this as: $49 \cdot 7^{-5} = 7^2 \cdot 7^{-5} = 7^{2-5} = 7^{-3} = \frac{1}{7^3}.$

Step 3: Combine these results: $45 \cdot \frac{1}{7^3} = \frac{45}{7^3}.$

Therefore, the solution to the problem is $\frac{45}{7^3}$ .

Final Answer

$\frac{45}{7^3}$

Key Points to Remember

Essential concepts to master this topic

Rule: $a^{-m} = \frac{1}{a^m}$ and $a^m \cdot a^n = a^{m+n}$
Technique: Convert $\frac{1}{45^{-81}}$ to $45^{81}$ , then add exponents: -80 + 81 = 1
Check: Substitute back: $45 \cdot \frac{1}{7^3} = \frac{45}{343}$ matches our final answer ✓

Common Mistakes

Avoid these frequent errors

Adding exponents incorrectly when dividing by negative powers
Don't treat $\frac{1}{45^{-81}}$ as division = wrong sign! This leads to $45^{-80-81} = 45^{-161}$ instead of the correct result. Always convert negative exponents first: $\frac{1}{a^{-n}} = a^n$ , then add exponents properly.

Practice Quiz

Test your knowledge with interactive questions

$112^0=\text{?}$

FAQ

Everything you need to know about this question

Why does $\frac{1}{45^{-81}}$ become $45^{81}$ ?

When you have a negative exponent in the denominator, it flips to become positive in the numerator! Think of it as: $\frac{1}{a^{-n}} = \frac{1}{\frac{1}{a^n}} = a^n$ .

How do I know when to add or subtract exponents?

Multiplying same bases: add exponents ( $a^m \cdot a^n = a^{m+n}$ ). Dividing same bases: subtract exponents ( $\frac{a^m}{a^n} = a^{m-n}$ ). In this problem, we're multiplying!

Why does 49 become $7^2$ ?

Because $7 \times 7 = 49$ , so $49 = 7^2$ . Converting to the same base (base 7) lets us use exponent rules: $7^2 \cdot 7^{-5} = 7^{2-5} = 7^{-3}$ .

What if I can't recognize that 49 is $7^2$ ?

Practice memorizing perfect squares! Common ones: $4=2^2, 9=3^2, 16=4^2, 25=5^2, 36=6^2, 49=7^2, 64=8^2, 81=9^2$ . These patterns appear frequently in exponent problems.

Can I leave my answer as $\frac{45}{7^3}$ or do I need to calculate $7^3$ ?

Both forms are correct! $\frac{45}{7^3}$ is the exact form, while $\frac{45}{343}$ is the decimal form. Check what your teacher prefers - often exact form is better.

How do I check if my final answer is right?

Substitute back into the original expression! Calculate each part: $45^{-80} \cdot 45^{81} = 45$ and $49 \cdot 7^{-5} = \frac{1}{7^3}$ , then multiply: $45 \cdot \frac{1}{7^3} = \frac{45}{7^3}$ ✓

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Solve the Exponent Equation: 45^-80 × 1/45^-81 × 49 × 7^-5

Exponent Rules with Negative Powers

❤️ 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 does $\frac{1}{45^{-81}}$ become $45^{81}$ ?

How do I know when to add or subtract exponents?

Why does 49 become $7^2$ ?

What if I can't recognize that 49 is $7^2$ ?

Can I leave my answer as $\frac{45}{7^3}$ or do I need to calculate $7^3$ ?

How do I check if my final answer is right?

🌟 Unlock Your Math Potential

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Solve the Exponent Equation: 45^-80 × 1/45^-81 × 49 × 7^-5

Exponent Rules with Negative Powers

❤️ 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 does 145−81 \frac{1}{45^{-81}} 45−811​ become 4581 45^{81} 4581?

How do I know when to add or subtract exponents?

Why does 49 become 72 7^2 72?

What if I can't recognize that 49 is 72 7^2 72?

Can I leave my answer as 4573 \frac{45}{7^3} 7345​ or do I need to calculate 73 7^3 73?

How do I check if my final answer is right?

🌟 Unlock Your Math Potential

More Questions

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Solve (2×4×8)^(a+3): Understanding Compound Power Expressions

Solve: 10^8 + 10^-4 + (1/10)^-16 Expression with Mixed Powers

Calculate the Result of the Fractions and Powers: 1/(-3)·3^(-4)·5^3

Evaluate the Expression: Finding Solutions for ((8by)^3)^y + (3^x)^a

Negative Exponents

Solve: (7^8)/(7^-4 × 4^2) × 32 Using Exponent Rules

Solve the Mixed Fraction and Fourth Root Equation: 2^3/3^2 * 3^-2 * Fourth Root of 81 = ?

Simplify the Expression: 7² × (3⁵)⁻¹ × (1/4) × (1/3²)

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Simplify Base-9 Expression: 9^300 × 1/9^-252 × 9^-549

Why does $\frac{1}{45^{-81}}$ become $45^{81}$ ?

Why does 49 become $7^2$ ?

What if I can't recognize that 49 is $7^2$ ?

Can I leave my answer as $\frac{45}{7^3}$ or do I need to calculate $7^3$ ?