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revisiting exponents & their functions quick check esmeralda wants to s…

Question

revisiting exponents & their functions quick check
esmeralda wants to solve for x in the equation $3^{-2x + 1} \cdot 3^{-2x - 3} = 3^{-3}$. which of the following answers should she select? (1 point)
\\( \circ \\ x = \frac{1}{4} \\)
\\( \circ \\ x = \frac{-5 \pm \sqrt{73}}{8} \\)
\\( \circ \\ x = -2 \\)
\\( \circ \\ x = -4 \\)

Explanation:

Step1: Use exponent rule \(a^m \cdot a^n = a^{m + n}\)

For the left - hand side of the equation \(3^{-2x + 1}\cdot3^{-2x-3}\), by the rule of exponents \(a^m\cdot a^n=a^{m + n}\), we have \(3^{-2x + 1+( - 2x-3)}=3^{-4x - 2}\)

Step2: Set exponents equal (since bases are equal)

The original equation is \(3^{-2x + 1}\cdot3^{-2x-3}=3^{-3}\), after simplifying the left - hand side, we get \(3^{-4x - 2}=3^{-3}\). Since the bases (\(a = 3\)) are the same and \(a^m=a^n\) implies \(m = n\) (for \(a>0,a
eq1\)), we can set the exponents equal: \(-4x-2=-3\)

Step3: Solve for x

Add 2 to both sides of the equation \(-4x-2=-3\): \(-4x=-3 + 2=-1\)
Then divide both sides by \(-4\): \(x=\frac{-1}{-4}=\frac{1}{4}\)

Answer:

\(x=\frac{1}{4}\) (the option \(x = \frac{1}{4}\))