What is the value of A when we rewrite 8^(x) as A^(-(x)/(9)) ? Choose 1 answer: (A) A=8^(-9) (B) A=-(9)/(8) (C) A=8^((1)/(9)) (D) A=(-8)^((1)/(9))

Expressing 8^x in the form of A^(-x/9): We need to express $ 8^x $ in the form of $ A^{-\frac{x}{9}} $. To do this, we need to find a value for $ A $ such that when raised to the power of $ -\frac{x}{9} $, it is equivalent to $ 8^x $.Rewriting 8^x as (8^(1/9))^(9x): First, let's rewrite $ 8^x $ as $ (8^{\frac{1}{9}})^{9x} $ because $ 8^x = (8^{\frac{1}{9}})^{9x} $. This is because raising a number to a power and then raising it to another power is the same as multiplying the exponents.Taking the reciprocal of the base: Now, we want to match the form $ A^{-\frac{x}{9}} $. To do this, we need to take the reciprocal of the base inside the parentheses, which will change the sign of the exponent. So, $ (8^{\frac{1}{9}})^{9x} $ can be written as $ (8^{-\frac{1}{9}})^{-9x} $.Determining the value of A: We can now see that $ A $ must be equal to $ 8^{-\frac{1}{9}} $ because $ (8^{-\frac{1}{9}})^{-9x} $ is in the form of $ A^{-\frac{x}{9}} $ with $ A $ being $ 8^{-\frac{1}{9}} $.Correct answer: A = 8^(1/9): Therefore, the correct answer is $ A = 8^{\frac{1}{9}} $, which corresponds to choice (C).

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What is the value of
A when we rewrite
8^(x) as
A^(-(x)/(9)) ?
Choose 1 answer:
(A)
A=8^(-9)
(B)
A=-(9)/(8)
(C)
A=8^((1)/(9))
(D)
A=(-8)^((1)/(9))

What is the value of $A$ when we rewrite $8^{x}$ as $A^{-\frac{x}{9}}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $A=8^{-9}$ $\newline$ (B) $A=-\frac{9}{8}$ $\newline$ (C) $A=8^{\frac{1}{9}}$ $\newline$ (D) $A=(-8)^{\frac{1}{9}}$

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Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

Q. What is the value of

A

when we rewrite

8^{x}

A^{-\frac{x}{9}}

\newline

Choose

1

answer:

\newline

(A)

A=8^{-9}

\newline

(B)

A=-\frac{9}{8}

\newline

(C)

A=8^{\frac{1}{9}}

\newline

(D)

A=(-8)^{\frac{1}{9}}

Expressing $8$ ^x in the form of A^(-x/ $9$ ): We need to express $8^x$ in the form of $A^{-\frac{x}{9}}$ . To do this, we need to find a value for $A$ such that when raised to the power of $-\frac{x}{9}$ , it is equivalent to $8^x$ .
Rewriting $8$ ^x as ( $8$ ^( $1$ / $9$ ))^( $9$ x): First, let's rewrite $8^x$ as $(8^{\frac{1}{9}})^{9x}$ because $8^x = (8^{\frac{1}{9}})^{9x}$ . This is because raising a number to a power and then raising it to another power is the same as multiplying the exponents.
Taking the reciprocal of the base: Now, we want to match the form $A^{-\frac{x}{9}}$ . To do this, we need to take the reciprocal of the base inside the parentheses, which will change the sign of the exponent. So, $(8^{\frac{1}{9}})^{9x}$ can be written as $(8^{-\frac{1}{9}})^{-9x}$ .
Determining the value of A: We can now see that $A$ must be equal to $8^{-\frac{1}{9}}$ because $(8^{-\frac{1}{9}})^{-9x}$ is in the form of $A^{-\frac{x}{9}}$ with $A$ being $8^{-\frac{1}{9}}$ .
Correct answer: A = $8$ ^( $1$ / $9$ ): Therefore, the correct answer is $A = 8^{\frac{1}{9}}$ , which corresponds to choice (C).