Question

part i questions: answer all questions in this part by writing the choice of the appropriate answer in the blank beside the problem. each question is worth 2 points. no partial credit will be awarded.

1. which of the following is not equivalent to the expression $5^{-2x}$?

(1) $\left(\frac{1}{25}\
ight)^x$ (3) $(-25)^x$
(2) $\left(5^{-2}\
ight)^x$ (4) $\left(5^x\
ight)^{-2}$

1. the expression $m^{\frac{3}{5}}$ can be rewritten equivalently as which of the following?

(1) $5\sqrt{m^3}$
(2) $\sqrt3{m^5}$
(3) $3\sqrt{m}$
(4) $\sqrt5{m^3}$

1. if $h(x) = 20(1.25)^x$, then which of the following statements describes the end behavior of this function?

(1) as $x \to -\infty$, $h(x) \to 0$ and as $x \to \infty$, $h(x) \to \infty$
(2) as $x \to -\infty$, $h(x) \to \infty$ and as $x \to \infty$, $h(x) \to 0$
(3) as $x \to -\infty$, $h(x) \to 20$ and as $x \to \infty$, $h(x) \to \infty$
(4) as $x \to -\infty$, $h(x) \to 0$ and as $x \to \infty$, $h(x) \to 1.25$

1. an exponential function of the form $y = a(b)^x$ passes through the points $(2, 35)$ and $(6, 10)$. which of the following is closest to the value of $b$ for this function?

(1) 1.28
(2) 1.09
(3) 0.73
(4) 0.62

1. which of the following is the solution to $8^{2x - 5} = \frac{1}{32}$

(1) $\frac{5}{2}$ (3) $\frac{8}{3}$
(2) $\frac{5}{3}$ (4) $\frac{1}{2}$

Explanation:

Question 1

Step1: Recall exponent rules

We know that \(a^{mn}=(a^m)^n=(a^n)^m\) and \(a^{-n}=\frac{1}{a^n}\). Let's simplify each option:

• Option (1): \(5^{-2x}=(5^{-2})^x = (\frac{1}{25})^x\), so this is equivalent.
• Option (2): By the power - of - a - power rule \((a^m)^n=a^{mn}\), \((5^{-2})^x = 5^{-2x}\), so this is equivalent.
• Option (3): \((-25)^x=(-1\times25)^x=(-1)^x\times25^x\), and \(5^{-2x}=\frac{1}{25^x}\) (when \(x\) is an integer, the signs and the bases are different), so this is not equivalent.
• Option (4): By the power - of - a - power rule \((a^m)^n=a^{mn}\), \((5^{x})^{-2}=5^{-2x}\), so this is equivalent.

Step1: Recall the rule for fractional exponents

The rule for fractional exponents is \(a^{\frac{m}{n}}=\sqrt[n]{a^m}\). For the expression \(m^{\frac{3}{5}}\), using the rule \(a^{\frac{m}{n}}=\sqrt[n]{a^m}\), we have \(m^{\frac{3}{5}}=\sqrt[5]{m^3}\).

Let's check the other options:

• Option (1): \(5\sqrt{m^3}=5m^{\frac{3}{2}}

eq m^{\frac{3}{5}}\)

• Option (2): \(\sqrt[3]{m^5}=m^{\frac{5}{3}}

eq m^{\frac{3}{5}}\)

• Option (3): \(3\sqrt{m}=3m^{\frac{1}{2}}

eq m^{\frac{3}{5}}\)

Step1: Analyze the exponential function \(h(x) = 20(1.25)^x\)

The general form of an exponential function is \(y = ab^x\), where \(a>0\) and \(b > 1\) represents exponential growth, and \(0 < b<1\) represents exponential decay. Here, \(a = 20\) and \(b=1.25>1\), so the function is an exponential growth function.

• As \(x

ightarrow-\infty\): We know that for \(y = ab^x\) with \(b>1\), when \(x
ightarrow-\infty\), \(b^x=\frac{1}{b^{\vert x\vert}}
ightarrow0\) (since \(b^{\vert x\vert}
ightarrow\infty\) as \(\vert x\vert
ightarrow\infty\)). So \(h(x)=20\times b^x
ightarrow20\times0 = 0\) as \(x
ightarrow-\infty\).

• As \(x

ightarrow\infty\): Since \(b = 1.25>1\), \(b^x
ightarrow\infty\) as \(x
ightarrow\infty\), so \(h(x)=20\times b^x
ightarrow\infty\) as \(x
ightarrow\infty\).

So the statement in option (1) is correct.

Answer:

(3) \((-25)^x\)

Question 2