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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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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}

as

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 \(48\).

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