Question

Simple question:

Answer 1

Does \[k^25^{-k^{3}}=\frac{ k^2 }{ 5^{-k^{3}} }\]

or

\[\frac{ k^2 }{ 5k^3 }\]

Answer 2

It's just a step im using in evaluating a improper integral and I forgot the exponent rules here

Answer 3

It's the latter that is correct.

Answer 4

I know \[(x^a)^b=x^{a*b}\]

Answer 5

could you explain?

Answer 6

Example:\[a ^{-1}=\frac{ 1 }{ a }\]

Answer 7

\[k^25^{-k^{3}}=\frac{ k^2 }{ 5^{-k^{3}} }\]

Answer 8

...is incorrect because you kept the negative sign in \[5^{-k^2}\]

Answer 9

whoops, didnt see that

Answer 10

take out the negative sign in your 1st option to make it correct

Answer 11

If you re-write this as \[\frac{ 1 }{ 5^{k^2} }\]

Answer 12

your original expression equals this result.

Answer 13

What is the simplified version of -7y^(-2) ?

Answer 14

\(5k^3 \neq 5^{k^{3}}\) though

Answer 15

The inequality symbol you used here is correct; the left side does not equal the right side.

Answer 16

\[\frac{ 7 }{ y^2 }\]

Answer 17

The focus of this problem is what to do when you have a negative exponent and wish to simplify the expression by writing an equivalent expression with a positive exponent.

Answer 18

\(\color{blue}{\text{Originally Posted by}}\) @mathmale
It's the latter that is correct.
\(\color{blue}{\text{End of Quote}}\)

Answer 19

Note that there are 2 negative signs in -7y^(-2).
Try again to simplify this. Goal: eliminate the negative exponent.

Answer 20

\[-7y^{-2} = \frac{ -7 }{ y^2 }\]

Answer 21

Very good. You have eliminated the neg exponent properly, and you show recognition that the initial coefficient, -7, does not change.

Answer 22

Im still unsure how to solve the original problem though. I understand the rule here, but I still get \(\frac{ k^2 }{ 5^{k^{3}} }\) when I solve

Answer 23

\[\large\rm \int\limits \frac{\color{royalblue}{k^2~dk}}{5^{\color{orangered}{k^3}}}\]Not seeing it?
Should be a nice u-sub you can apply.

Answer 24

\[5^{-k^{3}}=\frac{ 1 }{ 5^{k^{3}} } = \frac{ 1 }{ 5^{3k} }\]

Answer 25

No, that last step is no beuno.

Answer 26

\[\large\rm a^{b^c}\ne a^{bc}\]
\[\large\rm (a^b)^c=a^{bc}\]

Answer 27

\[(k^a)^b=k^{a*b} no?\]

Answer 28

ohh whoops

Answer 29

The 3 is not being applied to the entire "thing",
so you cannot apply that rule.

Answer 30

Still though, how does this simplify down to 5k^3?

Answer 31

Honestly, you can just do u-sub from the start if you prefer.\[\large\rm \int\limits 5^{-k^3}k^2~dk\]
\[\large\rm u=-k^3\]

Answer 32

It doesn't. Mathmale must've misread your original statement.

Answer 33

OHHH
ha, thats why I was confused

Answer 34

One of my classmates made the same mistake, so I was convinced it was right

Answer 35

Ahh what a bozo :D

Answer 36

no worries I called him out on it lol