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$Ms. Kent writes the two functions shown in the box on the whiteboard and asks her students to solve the system. {:[f(x)=0.25(0.5)^(x)],[g(x)=(2)^(x)]:} Which is the solution to this system? (A) x=-1 (B) x=0 (C) x=0.5 (D) x=1$

Ms. Kent writes the two functions shown in the box on the whiteboard and asks her students to solve the system. $\newline$ $\begin{array}{l}f(x)=0.25(0.5)^{x} \\ g(x)=(2)^{x}\end{array}$ $\newline$ Which is the solution to this system? $\newline$ (A) $x=-1$ $\newline$ (B) $x=0$ $\newline$ (C) $x=0.5$ $\newline$ (D) $x=1$

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Math Problems

Precalculus

Solve quadratic equations: word problems

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Q. Ms. Kent writes the two functions shown in the box on the whiteboard and asks her students to solve the system.

\newline

\begin{array}{l}f(x)=0.25(0.5)^{x} \\ g(x)=(2)^{x}\end{array}

\newline

Which is the solution to this system?

\newline

(A)

x=-1

\newline

(B)

x=0

\newline

(C)

x=0.5

\newline

(D)

x=1

Set Equal Functions: Set the two functions equal to each other to find the point of intersection. $\newline$ $f(x) = g(x)$ $\newline$ $0.25(0.5)^x = (2)^x$
Rewrite with Exponents: Recognize that $0.5$ is the same as $2^{-1}$ and rewrite the equation. $\newline$ $0.25 \times (2^{-1})^x = 2^x$
Simplify Left Side: Simplify the left side of the equation using the property of exponents $a^{bc} = (a^b)^c$ . $0.25 \times 2^{-x} = 2^x$
Substitute Values: Recognize that $0.25$ is the same as $2^{-2}$ and substitute. $\newline$ $2^{-2} \times 2^{-x} = (2)^x$
Combine Exponents: Combine the exponents on the left side using the property of exponents $a^{b+c} = a^b \cdot a^c$ . $2^{-2-x} = (2)^x$
Set Bases Equal: Since the bases are the same, set the exponents equal to each other. $\newline$ $-2 - x = x$
Solve for x: Solve for x by adding $x$ to both sides. $\newline$ $-2 = 2x$
Isolate $x$ : Divide both sides by $2$ to isolate $x$ . $\newline$ $x = \frac{-2}{2}$ $\newline$ $x = -1$