What would this be using a positive exponent?

Mathematics

1 answer:

tatuchka [14]3 years ago

5 0

1 / 9^4 and 1 / 6^3

This is because in order to make a negative exponent into a postive exponent, you need to make it into a fraction, 1 as the top number, and the negative exponent as a postive one.

For example, 6^-3 becomes 1 / 6^3.

I hope I helped.

You might be interested in

1.Simplify 5 to the 5th over 5 to the 8th.

timurjin [86]

Q1. The answer is 1 over 5 to the 3rd power.

5 to the 5th is 5⁵
5 to the 8th is 5⁸
5 to the 5th over 5 to the 8th is $\frac{ 5^{5} }{ 5^{8} }$
To simplify, we will use two rules:
$\frac{ x^{a}}{ x^{b} } = x^{a-b} \\ x^{-a} = \frac{1}{ x^{a}}$
Therefore:
$\frac{ 5^{5} }{ 5^{8} } = 5^{5-8} =5^{-3}= \frac{1}{5^{3}}$
$\frac{1}{5^{3}}$ is the same as 1 over 5 to the 3rd power.

Q2. The answer is (53)−4.

The exponents are multiplied when are expressed in the form:
 $(x^{a}) ^{b}= x^{a*b}$
So, $(5^{3} )^{-4}=5^{3*(-4)}=5^{-12}$

Other choices are incorrect:
one third to the 4th times one third to the 7th is $( \frac{1}{3} )^{4} *( \frac{1}{3} )^{7}=( \frac{1}{3} )^{4+7}$
3 to the 7th over 3 to the 15th is $\frac{ 3^{7} }{3^{15}} =3^{7-15}$
45 ⋅ 42 is $4^{5}* 4^{2} = 4^{5+2}$