Question

help #2

Answer 1

@waterineyes

Answer 2

\[\large (t^p)^q = (t)^{p \times q}\]

Answer 3

So:
\[\large (a^{-2})^{-1} = (a)^{-2 \times -1} = (a)^2\]

Answer 4

Similarly, can you solve for following:
\[\large (ax^3)^{-5} =??\]

Answer 5

And one thing:
\[(pt)^q = p^q \cdot t^q\]

Answer 6

\[1/a^5/ x^\] 15

Answer 7

It is -5 no?

Answer 8

1*(-5) = -5
3*(-5) = -15

Answer 9

\[(ax^3)^{-5} = a^{-5} \cdot x^{-15}\]

Answer 10

Nope..

Answer 11

two or three more steps to go now..

Answer 12

One more:
\[\frac{t^p}{t^q} = t^{p-q}\]

Answer 13

\[\frac{a^2}{(a^{-5}) \cdot (x)^{-15}}\]

Answer 14

\[\frac{a^{2-(-5)}}{x^{-15}} = \frac{a^7}{x^{-15}}\]

Answer 15

One more:
\[\frac{1}{t^{-p}} = t^p\]

Answer 16

So, finally you will get:
\[a^7 \cdot x^{15}\]