What is the value of A when we rewrite ((2)/(3))^(x+4)-((2)/(3))^(x) as A*((2)/(3))^(x) ? A=

Factor out common term: We need to factor out ((2)/(3))^x from both terms in the expression ((2)/(3))^(x+4)-((2)/(3))^(x).Rewrite using exponent property: First, we rewrite ((2)/(3))^(x+4) as ((2)/(3))^x * ((2)/(3))^4 by using the property of exponents that states a^(m+n) = a^m * a^n.Factor out common term again: Now we have ((2)/(3))^x * ((2)/(3))^4 - ((2)/(3))^x. We can factor ((2)/(3))^x out of both terms.Calculate value inside brackets: After factoring out ((2)/(3))^x, we get ((2)/(3))^x * [((2)/(3))^4 - 1].Substitute value into brackets: We calculate ((2)/(3))^4 to find the value inside the brackets. ((2)/(3))^4 = (2^4)/(3^4) = 16/81.Calculate final result: Substitute 16/81 into the brackets: ((2)/(3))^x * (16/81 - 1).Now we calculate 16/81 - 1. Since 1 is equivalent to 81/81, we have 16/81 - 81/81 = (16 - 81)/81 = -65/81.The expression is now ((2)/(3))^x * (-65/81). Therefore, A = -65/81.

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What is the value of
A when we rewrite
((2)/(3))^(x+4)-((2)/(3))^(x) as
A*((2)/(3))^(x) ?

A=

What is the value of $A$ when we rewrite $\left(\frac{2}{3}\right)^{x+4}-\left(\frac{2}{3}\right)^{x}$ as $A \cdot\left(\frac{2}{3}\right)^{x}$ ? $\newline$ $A=$

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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{2}{3}\right)^{x+4}-\left(\frac{2}{3}\right)^{x}

A \cdot\left(\frac{2}{3}\right)^{x}

\newline

A=

Factor out common term: We need to factor out $\left(\frac{2}{3}\right)^x$ from both terms in the expression $\left(\frac{2}{3}\right)^{x+4}-\left(\frac{2}{3}\right)^{x}$ .
Rewrite using exponent property: First, we rewrite $\left(\frac{2}{3}\right)^{x+4}$ as $\left(\frac{2}{3}\right)^x \cdot \left(\frac{2}{3}\right)^4$ by using the property of exponents that states $a^{m+n} = a^m \cdot a^n$ .
Factor out common term again: Now we have $\left(\frac{2}{3}\right)^x \cdot \left(\frac{2}{3}\right)^4 - \left(\frac{2}{3}\right)^x$ . We can factor $\left(\frac{2}{3}\right)^x$ out of both terms.
Calculate value inside brackets: After factoring out $\left(\frac{2}{3}\right)^x$ , we get $\left(\frac{2}{3}\right)^x \cdot \left[\left(\frac{2}{3}\right)^4 - 1\right].$
Substitute value into brackets: We calculate $\left(\frac{2}{3}\right)^4$ to find the value inside the brackets. $\left(\frac{2}{3}\right)^4 = \left(2^4\right)/\left(3^4\right) = \frac{16}{81}$ .
Calculate final result: Substitute $\frac{16}{81}$ into the brackets: $\left(\frac{2}{3}\right)^x \times \left(\frac{16}{81} - 1\right)$ .
Calculate final result: Substitute $\frac{16}{81}$ into the brackets: $\left(\frac{2}{3}\right)^x \times \left(\frac{16}{81} - 1\right)$ .Now we calculate $\frac{16}{81} - 1$ . Since $1$ is equivalent to $\frac{81}{81}$ , we have $\frac{16}{81} - \frac{81}{81} = \frac{16 - 81}{81} = -\frac{65}{81}.$
Calculate final result: Substitute $\frac{16}{81}$ into the brackets: $\left(\frac{2}{3}\right)^x \times \left(\frac{16}{81} - 1\right)$ .Now we calculate $\frac{16}{81} - 1$ . Since $1$ is equivalent to $\frac{81}{81}$ , we have $\frac{16}{81} - \frac{81}{81} = \frac{16 - 81}{81} = -\frac{65}{81}$ .The expression is now $\left(\frac{2}{3}\right)^x \times \left(-\frac{65}{81}\right)$ . Therefore, $A = -\frac{65}{81}$ .