Exponential Expressions Simplified

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Cindy Burgess

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Apr 13, 2013
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I just want to check my work before turning it in. These are the ones I wasn't sure about

1. (2 x 10^-8)(8 x 10^-10) = 1.6 x 10^-17

2. (6k^5/3 x 2j^1/4)(k^1/3 x 3j^1/8) = 36k^2j^3/8

3. (-4a^2b^3)^2 (a^5b^3)^3=16a^19b^15

4. a^-12b^3/a^-22b^13=a^10/b^10

5. (1/2n^6)^2 = 1/4n^12

6. (t^3/7c^2)^-4 = 2401c^8/t^12
 
Cindy Burgess said:
I just want to check my work before turning it in. These are the ones I wasn't sure about

1. (2 x 10^-8)(8 x 10^-10) = 1.6 x 10^-17............................................Correct

2. (6k^5/3 x 2j^1/4)(k^1/3 x 3j^1/8) = 36k^2j^3/8............................Incorrect

3. (-4a^2b^3)^2 (a^5b^3)^3=16a^19b^15............................................Correct

4. a^-12b^3/a^-22b^13=a^10/b^10............................Incorrect (as posted)

5. (1/2n^6)^2 = 1/4n^12............................................Correct

6. (t^3/7c^2)^-4 = 2401c^8/t^12............................................Correct
Click to expand...

.
 
Last edited by a moderator:
Hello, Cindy!

If I read the problems correctly,
. . your answers to 2 and 4 are also correct.

2. (6k^5/3 x 2j^1/4)(k^1/3 x 3j^1/8) = 36k^2j^3/8
Click to expand...

\(\displaystyle \left(6k^{\frac{5}{3}}\cdot2j^{\frac{1}{4}} \right)\left(k^{\frac{1}{3}}\cdot3j^{\frac{1}{8}} \right) \;=\;(6\cdot2\cdot3)\left(k^{\frac{5}{3}}\cdot k^{\frac{1}{3}}\right)\left(j^{\frac{1}{4}}\cdot j^{\frac{1}{8}}\right) \;=\;36k^2j^{\frac{3}{8}}\)



4. a^-12b^3/a^-22b^13 = a^10/b^10
Click to expand...

This one needs parentheses and x's: .(a^-12 x b^3) / (a^-22 x b^13)

. . \(\displaystyle \dfrac{a^{\text{-}12}b^3}{a^{\text{-}22}b^{13}} \;=\;\dfrac{a^{\text{-}12-(\text{-}22)}}{b^{13-3}} \;=\; \dfrac{a^{10}}{b^{10}} \)
 
Thank you. You wrote them correctly. I appreciate your help.

Hello, Cindy!

If I read the problems correctly,
. . your answers to 2 and 4 are also correct.


\(\displaystyle \left(6k^{\frac{5}{3}}\cdot2j^{\frac{1}{4}} \right)\left(k^{\frac{1}{3}}\cdot3j^{\frac{1}{8}} \right) \;=\;(6\cdot2\cdot3)\left(k^{\frac{5}{3}}\cdot k^{\frac{1}{3}}\right)\left(j^{\frac{1}{4}}\cdot j^{\frac{1}{8}}\right) \;=\;36k^2j^{\frac{3}{8}}\)




This one needs parentheses and x's: .(a^-12 x b^3) / (a^-22 x b^13)

. . \(\displaystyle \dfrac{a^{\text{-}12}b^3}{a^{\text{-}22}b^{13}} \;=\;\dfrac{a^{\text{-}12-(\text{-}22)}}{b^{13-3}} \;=\; \dfrac{a^{10}}{b^{10}} \)[/QUOTE]
 
Cindy Burgess said:
I just want to check my work before turning it in. These are the ones I wasn't sure about

1. (2 x 10^-8)(8 x 10^-10) = 1.6 x 10^-17

2. (6k^5/3 x 2j^1/4)(k^1/3 x 3j^1/8) = 36k^2j^3/8

3. (-4a^2b^3)^2 (a^5b^3)^3=16a^19b^15

4. a^-12b^3/a^-22b^13=a^10/b^10

5. (1/2n^6)^2 = 1/4n^12

6. (t^3/7c^2)^-4 = 2401c^8/t^12
Click to expand...
Cindy Burgess, when you type the problems and/or their solutions without certain needed grouping symbols, then they are wrong. In particular, you must have the grouping symbols placed around the fractional exponents. I tried to make to make necessary and suggested amendments below:

1. [2 x 10^(-8)][8 x 10^(-10)] = 1.6 x 10^(-17)

2. [6k^(5/3) x 2j^(1/4)][k^(1/3) x 3j^(1/8)] = 36k^2j^(3/8)

3. [4a^2b^3)^2][(a^5b^3)^3] = 16a^19b^15

4. [a^(-12)b^3]/[a^(-22)b^13] = a^10/b^10

5. [1/(2n^6)]^2 = 1/(4n^12)

6. [t^3/(7c^2)]^(-4) = 2401c^8/t^12
 
Moreover, use '*' to indicate 'multiplication' or 'times' - instead of 'x' or '.'
 
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