What is the value of A when we rewrite ((1)/(8))^(x)+((1)/(8))^(x-2) as A*((1)/(8))^(x) ? A=

Express with Same Exponent: First, we need to express both terms with the same base and exponent. The first term is already in the form we want, ((1)/(8))^x. We need to manipulate the second term, ((1)/(8))^(x-2), to have the same exponent as the first term.Apply Exponent Property: To do this, we can use the property of exponents that states a^(m-n) = a^m / a^n. We apply this to the second term to get ((1)/(8))^(x) / ((1)/(8))^2.Rewrite Second Term: Since ((1)/(8))^2 is equal to (1/64), we can rewrite the second term as ((1)/(8))^x * (1/(1/64)) or ((1)/(8))^x * 64.Factor Out Common Term: Now we have the expression ((1)/(8))^x + 64*((1)/(8))^x. We can factor out ((1)/(8))^x to get ((1)/(8))^x * (1 + 64).Calculate Final Value: Adding 1 + 64 gives us 65. So the expression becomes 65 * ((1)/(8))^x.Therefore, the value of A when we rewrite ((1)/(8))^x + ((1)/(8))^(x-2) as A*((1)/(8))^x is 65.

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What is the value of
A when we rewrite
((1)/(8))^(x)+((1)/(8))^(x-2) as
A*((1)/(8))^(x) ?

A=

What is the value of $A$ when we rewrite $\left(\frac{1}{8}\right)^{x}+\left(\frac{1}{8}\right)^{x-2}$ as $A \cdot\left(\frac{1}{8}\right)^{x}$ ? $\newline$ $A=$

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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}{8}\right)^{x}+\left(\frac{1}{8}\right)^{x-2}

A \cdot\left(\frac{1}{8}\right)^{x}

\newline

A=

Express with Same Exponent: First, we need to express both terms with the same base and exponent. The first term is already in the form we want, $\left(\frac{1}{8}\right)^x$ . We need to manipulate the second term, $\left(\frac{1}{8}\right)^{x-2}$ , to have the same exponent as the first term.
Apply Exponent Property: To do this, we can use the property of exponents that states $a^{(m-n)} = \frac{a^m}{a^n}$ . We apply this to the second term to get $\left(\frac{1}{8}\right)^x / \left(\frac{1}{8}\right)^2$ .
Rewrite Second Term: Since $\left(\frac{1}{8}\right)^2$ is equal to $\frac{1}{64}$ , we can rewrite the second term as $\left(\frac{1}{8}\right)^x \cdot \left(\frac{1}{\frac{1}{64}}\right)$ or $\left(\frac{1}{8}\right)^x \cdot 64$ .
Factor Out Common Term: Now we have the expression $((1)/(8))^x + 64*((1)/(8))^x$ . We can factor out $((1)/(8))^x$ to get $((1)/(8))^x * (1 + 64)$ .
Calculate Final Value: Adding $1 + 64$ gives us $65$ . So the expression becomes $65 \times \left(\frac{1}{8}\right)^x$ .
Calculate Final Value: Adding $1 + 64$ gives us $65$ . So the expression becomes $65 \times \left(\frac{1}{8}\right)^x$ .Therefore, the value of $A$ when we rewrite $\left(\frac{1}{8}\right)^x + \left(\frac{1}{8}\right)^{x-2}$ as $A\times\left(\frac{1}{8}\right)^x$ is $65$ .