## anonymous one year ago What is the missing value? 3^-6/3^?=3^4 A. −10 B. −2 C. 2 D. 10

1. whpalmer4

$\frac{3^{-6}}{3^?} = 3^4$ Remember that $\frac{a^m}{a^n} = a^{m-n}$

2. whpalmer4

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

3. anonymous

???

4. whpalmer4

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.

5. whpalmer4

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.