Answer :
Certainly! Let's determine the values of \(a\) and \(b\) for the exponential function \(f(x) = a \cdot b^x\) given that it passes through the points \((0, 11)\) and \((3, 88)\).
### Step 1: Determine \(a\)
First, we use the point \((0, 11)\).
When \(x = 0\):
[tex]\[ f(0) = a \cdot b^0 \][/tex]
Since \(b^0 = 1\):
[tex]\[ f(0) = a \cdot 1 = a \][/tex]
Given that \(f(0) = 11\):
[tex]\[ a = 11 \][/tex]
So, we have:
[tex]\[ a = 11 \][/tex]
### Step 2: Determine \(b\)
Now, using the second point \((3, 88)\).
When \(x = 3\):
[tex]\[ f(3) = a \cdot b^3 \][/tex]
We already found that \(a = 11\), so:
[tex]\[ 88 = 11 \cdot b^3 \][/tex]
To solve for \(b\), divide both sides by 11:
[tex]\[ \frac{88}{11} = b^3 \][/tex]
[tex]\[ 8 = b^3 \][/tex]
Now, take the cube root of both sides to find \(b\):
[tex]\[ b = \sqrt[3]{8} \][/tex]
[tex]\[ b = 2 \][/tex]
### Final values:
[tex]\[ \begin{array}{l} a = 11 \\ b = 2 \end{array} \][/tex]
Thus, the values are:
[tex]\[ a = 11 \][/tex]
[tex]\[ b = 2 \][/tex]
### Step 1: Determine \(a\)
First, we use the point \((0, 11)\).
When \(x = 0\):
[tex]\[ f(0) = a \cdot b^0 \][/tex]
Since \(b^0 = 1\):
[tex]\[ f(0) = a \cdot 1 = a \][/tex]
Given that \(f(0) = 11\):
[tex]\[ a = 11 \][/tex]
So, we have:
[tex]\[ a = 11 \][/tex]
### Step 2: Determine \(b\)
Now, using the second point \((3, 88)\).
When \(x = 3\):
[tex]\[ f(3) = a \cdot b^3 \][/tex]
We already found that \(a = 11\), so:
[tex]\[ 88 = 11 \cdot b^3 \][/tex]
To solve for \(b\), divide both sides by 11:
[tex]\[ \frac{88}{11} = b^3 \][/tex]
[tex]\[ 8 = b^3 \][/tex]
Now, take the cube root of both sides to find \(b\):
[tex]\[ b = \sqrt[3]{8} \][/tex]
[tex]\[ b = 2 \][/tex]
### Final values:
[tex]\[ \begin{array}{l} a = 11 \\ b = 2 \end{array} \][/tex]
Thus, the values are:
[tex]\[ a = 11 \][/tex]
[tex]\[ b = 2 \][/tex]
