anonymous
  • anonymous
What is the missing value? 3^-6/3^?=3^4 A. −10 B. −2 C. 2 D. 10
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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whpalmer4
  • whpalmer4
\[\frac{3^{-6}}{3^?} = 3^4\] Remember that \[\frac{a^m}{a^n} = a^{m-n}\]
whpalmer4
  • whpalmer4
So that means \[\frac{3^{-6}}{3^?} = 3^{-6-?} = 3^4\]or \[-6-?=4\]Can you solve for ?
anonymous
  • anonymous
???

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whpalmer4
  • whpalmer4
The exponent needs to be a number such that -6 - = 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.
whpalmer4
  • 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.

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