Negative Exponents: Using the laws of exponents

Using variablesCalculating powers with negative exponentsComplete the equationIdentify the greater valueMultiplying Exponents with the same baseApplying the formulaconverting Negative Exponents to Positive ExponentsPresenting powers in the denominator as powers with negative exponentsPresenting powers with negative exponents as fractionsUsing laws of exponents with parameters
Question 1

\( (8\times9\times5\times3)^{-2}= \)

Question 2

\( (\frac{2}{3})^{-4}=\text{?} \)

Question 3

\( 7^2\cdot(3^5)^{-1}\cdot\frac{1}{4}\cdot\frac{1}{3^2}=\text{?} \)

Question 4

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

Question 5

\( 10^8+10^{-4}+(\frac{1}{10})^{-16}=\text{?} \)

Examples with solutions for Negative Exponents: Using the laws of exponents

Exercise #1

(8×9×5×3)−2= (8\times9\times5\times3)^{-2}=

Video Solution

Step-by-Step Solution

We begin by applying the power rule to the products within the parentheses:

(z⋅t)n=zn⋅tn (z\cdot t)^n=z^n\cdot t^n That is, the power applied to a product within parentheses is applied to each of the terms when the parentheses are opened,

We apply the rule to the given problem:

(8⋅9⋅5⋅3)−2=8−2⋅9−2⋅5−2⋅3−2 (8\cdot9\cdot5\cdot3)^{-2}=8^{-2}\cdot9^{-2}\cdot5^{-2}\cdot3^{-2} Therefore, the correct answer is option c.

Note:

Whilst it could be understood that the above power rule applies only to two terms of the product within parentheses, in reality, it is also valid for the power over a multiplication of multiple terms within parentheses, as was seen in the above problem.

A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms within parentheses (as formulated above), then it is also valid for a power over several terms of the product within parentheses (for example - three terms, etc.).

Answer

8−2×9−2×5−2×3−2 8^{-2}\times9^{-2}\times5^{-2}\times3^{-2}

Exercise #2

(23)−4=? (\frac{2}{3})^{-4}=\text{?}

Video Solution

Step-by-Step Solution

We use the formula:

(ab)−n=(ba)n (\frac{a}{b})^{-n}=(\frac{b}{a})^n

Therefore, we obtain:

(32)4 (\frac{3}{2})^4

We use the formula:

(ba)n=bnan (\frac{b}{a})^n=\frac{b^n}{a^n}

Therefore, we obtain:

3424=3×3×3×32×2×2×2=8116 \frac{3^4}{2^4}=\frac{3\times3\times3\times3}{2\times2\times2\times2}=\frac{81}{16}

Answer

8116 \frac{81}{16}

Exercise #3

72⋅(35)−1⋅14⋅132=? 7^2\cdot(3^5)^{-1}\cdot\frac{1}{4}\cdot\frac{1}{3^2}=\text{?}

Video Solution

Answer

4−13−77−2 \frac{4^{-1}3^{-7}}{7^{-2}}

Exercise #4

45−80⋅145−81⋅49⋅7−5=? 45^{-80}\cdot\frac{1}{45^{-81}}\cdot49\cdot7^{-5}=\text{?}

Video Solution

Answer

4573 \frac{45}{7^3}

Exercise #5

108+10−4+(110)−16=? 10^8+10^{-4}+(\frac{1}{10})^{-16}=\text{?}

Video Solution

Answer

108+1104+1016 10^8+\frac{1}{10^4}+10^{16}

Question 1

\( 3^x\cdot\frac{1}{3^{-x}}\cdot3^{2x}=\text{?} \)

Question 2

\( 5^4-(\frac{1}{5})^{-3}\cdot5^{-2}=\text{?} \)

Question 3

\( \frac{2^{-4}\cdot(\frac{1}{2})^8\cdot2^{10}}{2^3}=\text{?} \)

Question 4

\( 9^4\cdot3^{-8}\cdot\frac{1}{3}=\text{?} \)

Question 5

\( \frac{7^8}{7^{-4}\cdot4^2}\cdot32=\text{?} \)

Exercise #6

3x⋅13−x⋅32x=? 3^x\cdot\frac{1}{3^{-x}}\cdot3^{2x}=\text{?}

Video Solution

Answer

(34)x (3^4)^x

Exercise #7

54−(15)−3⋅5−2=? 5^4-(\frac{1}{5})^{-3}\cdot5^{-2}=\text{?}

Video Solution

Answer

5(53−1) 5(5^3-1)

Exercise #8

2−4⋅(12)8⋅21023=? \frac{2^{-4}\cdot(\frac{1}{2})^8\cdot2^{10}}{2^3}=\text{?}

Video Solution

Answer

2−5 2^{-5}

Exercise #9

94⋅3−8⋅13=? 9^4\cdot3^{-8}\cdot\frac{1}{3}=\text{?}

Video Solution

Answer

3−1 3^{-1}

Exercise #10

787−4⋅42⋅32=? \frac{7^8}{7^{-4}\cdot4^2}\cdot32=\text{?}

Video Solution

Answer

2⋅712 2\cdot7^{12}

Question 1

\( \frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{81}=\text{?} \)

Question 2

\( \frac{10^4\cdot0.1^{-3}\cdot10^{-8}}{1000}=\text{?} \)

Exercise #11

2332⋅3−2⋅814=? \frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{81}=\text{?}

Video Solution

Answer

(23)3 (\frac{2}{3})^3

Exercise #12

104⋅0.1−3⋅10−81000=? \frac{10^4\cdot0.1^{-3}\cdot10^{-8}}{1000}=\text{?}

Video Solution

Answer

0.0001 0.0001