Question

What is the missing value?

3^-6/3^?=3^4


A.
−10

B.
−2

C.
2

D.
10

Answer 1

\[\frac{3^{-6}}{3^?} = 3^4\]

Remember that
\[\frac{a^m}{a^n} = a^{m-n}\]

Answer 2

So that means
\[\frac{3^{-6}}{3^?} = 3^{-6-?} = 3^4\]or \[-6-?=4\]Can you solve for ?

Answer 3

???

Answer 4

The exponent needs to be a number such that -6 -<some number> = 4. Maybe you're confused because I didn't use a typical variable name, like \(x\).

\[-6-x = 4\]

Can you solve that for \(x\)? \(x\) is the exponent for the number in the denominator in the problem.

Answer 5

Or, you could solve it a different way:

\[\frac{3^{-6}}{3^x} = 3^4\]Let's multiply both sides by \(3^x\):

\[3^x*\frac{3^{-6}}{3^x} = 3^4*3^x\]\[\cancel{3^x*}\frac{3^{-6}}{\cancel{3^x}} = 3^4*3^x\]\[3^{-6} = 3^4*3^x\]

Now when we multiply exponentials with the same base, as we have here (everything is 3 to some power, 3 is the base), we keep the base and ADD the exponents.

\[3^2*3^3 = 3^{2+3} = 3^5\]just like if we did it the long way:\[3^2*3^3=(3*3)*(3*3*3) = 3*3*3*3*3=3^5\]

so \[3^{-6}=3^4*3^x \]\[3^{-6} = 3^{4+x}\]and the only way that can be true is if \[-6 = 4+x\]

Solve for \(x\) and you have your answer.