Question

What value of k solves the equation?



k^-3=1/27



A.
–81


B.
–9


C.
–3


D.
3



-3 and 3 have the same answers so I'm kind of stuck...

Answer 1

@jim_thompson5910

Answer 2

\(\large k^{-3} = \dfrac{1}{27} \)

Do you understand negative exponents?

Answer 3

For example, what is \(\large 5^{-2}\) ?

Answer 4

0.04?

Answer 5

Yes, that is correct, but my question is without using a calculator, how do you deal with a negative exponent?

Answer 6

In other words, what is \(\large 5^{-2} \) equal to using a positive exponent.

Answer 7

4%

Answer 8

Here it is. The general rule is:

\(\large a^{-n} = \dfrac{1}{a^n} \)

Here's the example I asked about above:

\( \large 5^{-2} = \dfrac{1}{5^2} \)

Answer 9

When you raise a number to a negative exponent, it means you write 1 over the number raised to the same but positive exponent.

Answer 10

Here are a few more examples:

\(\large 3^{-5} = \dfrac{1}{3^5} \)

\(\large x^{-6} = \dfrac{1}{x^6} \)

\(\large y^{-a} = \dfrac{1}{y^a} \)

Answer 11

Do you understand now what a negative exponent means?

Answer 12

Use this rule to change the left side of your equation to an expression with a positive exponent.

Answer 13

Is it 3?

Answer 14

Here is your equation again.
We need to follow the rule above and change just the left side to an expression with a positive exponent. Then we copy the right side just as it is.

\(\large k^{-3} = \dfrac{1}{27}\)

\( \large \dfrac{1}{k^3} = \dfrac{1}{27} \)

Answer 15

Isn't it 3?

Answer 16

You mean the final answer?

Answer 17

yes

Answer 18

\(\large \dfrac{1}{k^3} = \dfrac{1}{3^3} \)

\(k = 3\)

You are correct.

Answer 19

Thank You!

Answer 20

You're welcome.