What is the value of A when we rewrite ((1)/(7))^(x-2)-((1)/(7))^(x) as A*((1)/(7))^(x) ? A=

Factor out ((1)/(7))^(x): We need to express the given expression ((1)/(7))^(x-2) - ((1)/(7))^(x) in the form of A*((1)/(7))^(x). To do this, we will factor out ((1)/(7))^(x) from both terms.Rewrite ((1)/(7))^(x-2): First, let's rewrite ((1)/(7))^(x-2) by separating the exponent into two parts: ((1)/(7))^(x) * ((1)/(7))^(-2).Calculate ((1)/(7))^(-2): Now, we know that ((1)/(7))^(-2) is equal to 7^2, which is 49. So, ((1)/(7))^(x-2) can be rewritten as 49 * ((1)/(7))^(x).Substitute back into expression: Substitute this back into the original expression: 49 * ((1)/(7))^(x) - ((1)/(7))^(x).Factor out ((1)/(7))^(x): Now, factor out ((1)/(7))^(x) from both terms: ((1)/(7))^(x) * (49 - 1).Calculate expression inside parentheses: Calculate the expression inside the parentheses: 49 - 1 = 48.Final expression in desired form: We now have the expression in the desired form: A * ((1)/(7))^(x), where A is 48.

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What is the value of
A when we rewrite
((1)/(7))^(x-2)-((1)/(7))^(x) as
A*((1)/(7))^(x) ?

A=

What is the value of $A$ when we rewrite $\left(\frac{1}{7}\right)^{x-2}-\left(\frac{1}{7}\right)^{x}$ as $A \cdot\left(\frac{1}{7}\right)^{x}$ ? $\newline$ $A=$

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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

\left(\frac{1}{7}\right)^{x-2}-\left(\frac{1}{7}\right)^{x}

A \cdot\left(\frac{1}{7}\right)^{x}

\newline

A=

Factor out $(\frac{1}{7})^{x}$ : We need to express the given expression $(\frac{1}{7})^{(x-2)} - (\frac{1}{7})^{x}$ in the form of $A*(\frac{1}{7})^{x}$ . To do this, we will factor out $(\frac{1}{7})^{x}$ from both terms.
Rewrite $\left(\frac{1}{7}\right)^{x-2}$ : First, let's rewrite $\left(\frac{1}{7}\right)^{x-2}$ by separating the exponent into two parts: $\left(\frac{1}{7}\right)^{x} \cdot \left(\frac{1}{7}\right)^{-2}$ .
Calculate $\left(\frac{1}{7}\right)^{-2}$ : Now, we know that $\left(\frac{1}{7}\right)^{-2}$ is equal to $7^2$ , which is $49$ . So, $\left(\frac{1}{7}\right)^{x-2}$ can be rewritten as $49 \times \left(\frac{1}{7}\right)^{x}$ .
Substitute back into expression: Substitute this back into the original expression: $$49$ \times \left(\frac{$1$}{$7$}\right)^{x} - \left(\frac{$1$}{$7$}\right)^{x}.
Factor out $(\frac{1}{7})^{x}$: Now, factor out $(\frac{1}{7})^{x}$ from both terms: $(\frac{1}{7})^{x} \times (49 - 1)$.
Calculate expression inside parentheses: Calculate the expression inside the parentheses: $49 - 1 = 48$.
Final expression in desired form: We now have the expression in the desired form: $A \times \left(\frac{1}{7}\right)^{x}$, where $A$ is $48$.