Question

1. an increasing exponential function of the form $y = a(b)^x$ is shown below. algebraically determine the values of $a$ and $b$. round both values to the nearest thousandth. 4 points (graph with points (-2, 5) and (4, 80))
2. the expression $\frac{9^x}{243}$ can be written as $3^{f(x)}$, where $f(x)$ is some linear function. determine a formula for $f(x)$. justify your answer. 3 points
3. algebraically determine the $x$-coordinate where the functions $y = 5^{2x - 3}$ and $y = \left(\frac{1}{25}\

ight)^{x - 4}$ intersect. 3 points

Explanation:

Question 11

Step1: Substitute points into the function

We have the exponential function \( y = a(b)^x \) and two points \((-2, 5)\) and \((4, 80)\). Substitute \((-2, 5)\) into the function: \( 5=a(b)^{-2}=\frac{a}{b^{2}} \), so \( a = 5b^{2} \). Substitute \((4, 80)\) into the function: \( 80=a(b)^{4} \).

Step2: Substitute \( a = 5b^{2} \) into the second equation

Substitute \( a = 5b^{2} \) into \( 80=a(b)^{4} \), we get \( 80 = 5b^{2}\times b^{4}=5b^{6} \). Then \( b^{6}=\frac{80}{5} = 16 \), so \( b = 16^{\frac{1}{6}}=(2^{4})^{\frac{1}{6}}=2^{\frac{2}{3}}\approx1.587 \).

Step3: Find the value of \( a \)

Substitute \( b\approx1.587 \) into \( a = 5b^{2} \), we get \( a = 5\times(1.587)^{2}\approx5\times2.519\approx12.595 \).

Step1: Rewrite the bases

We know that \( 9 = 3^{2} \) and \( 243=3^{5} \). So the expression \( \frac{9^{x}}{243} \) can be rewritten as \( \frac{(3^{2})^{x}}{3^{5}} \).

Step2: Simplify the exponents

Using the power - of - a - power rule \((a^{m})^{n}=a^{mn}\) and the quotient rule \( \frac{a^{m}}{a^{n}}=a^{m - n} \), we have \( \frac{(3^{2})^{x}}{3^{5}}=\frac{3^{2x}}{3^{5}}=3^{2x - 5} \).
Since \( \frac{9^{x}}{243}=3^{f(x)} \), then \( f(x)=2x - 5 \).

Step1: Rewrite the second function's base

We know that \( \frac{1}{25}=5^{-2} \), so the function \( y = (\frac{1}{25})^{x - 4}=(5^{-2})^{x - 4} \).
Using the power - of - a - power rule \((a^{m})^{n}=a^{mn}\), we get \( (5^{-2})^{x - 4}=5^{-2(x - 4)}=5^{-2x + 8} \).

Step2: Set the two functions equal

Since the two functions \( y = 5^{2x-3} \) and \( y = 5^{-2x + 8} \) intersect, their exponents must be equal when their bases are equal (because the exponential function \( y = a^{x}(a>0,a
eq1) \) is one - to - one). So we set \( 2x-3=-2x + 8 \).

Step3: Solve for \( x \)

Add \( 2x \) to both sides of the equation \( 2x-3=-2x + 8 \): \( 2x+2x-3=-2x + 2x+8 \), which gives \( 4x-3 = 8 \).
Add 3 to both sides: \( 4x-3 + 3=8 + 3 \), so \( 4x=11 \).
Divide both sides by 4: \( x=\frac{11}{4}=2.75 \).

Answer:

\( a\approx12.595 \), \( b\approx1.587 \)

Question 12