Q. exponential function formula y=ab^x

Exponential function formula: y = a b^x — detailed step-by-step explanation

Below is a complete, careful explanation of the exponential function y = a b^x, how to interpret each parameter, how to determine a and b from data, how to solve for x or y, and important properties. Every algebraic step is shown and explained.

1. The form and meanings of the parameters

1. Definition: the general exponential function is written as

   \[ y = a\,b^x \]

   Here:

   • \(a\) is the initial factor (vertical scale). If \(x = 0\) then \(y = a\,b^0 = a\), so \(a\) is the value of the function at \(x=0\).
   • \(b\) is the base (growth/decay factor per unit increase in \(x\)). If \(b>1\) the function grows as \(x\) increases. If \(0

2. Domain and range:

   • Domain: all real numbers, \(x \in \mathbb{R}\), provided \(b>0\).
   • Range: if \(a\neq 0\) and \(b>0\), then \(y\) has the sign of \(a\) and cannot be zero (so range is either \((0,\infty)\) if \(a>0\), or \((-\infty,0)\) if \(a<0\)).

3. Restrictions: to have a standard exponential behavior we take \(b>0\) and usually \(b\neq 1\). If \(b=1\) the function is constant \(y=a\).

2. Determine a and b given information

Case A — Given the value at \(x=0\)

1. If you are given the point \((0,y_0)\) that lies on the graph, substitute \(x=0\) into the formula:
2. Substitution: \[ y_0 = a\,b^0 = a \times 1 = a. \]
3. Conclusion: \[ a = y_0. \] This determines \(a\) directly.

Case B — Given two points \((x_1,y_1)\) and \((x_2,y_2)\) with \(y_1\neq 0,\ y_2\neq 0\)

We can solve for \(a\) and \(b\) from two distinct points. Proceed step by step.

1. Write the two equations from the two points:

\[ y_1 = a\,b^{x_1} \]

\[ y_2 = a\,b^{x_2} \]

2. Divide the second equation by the first to eliminate \(a\):

\[ \frac{y_2}{y_1} = \frac{a\,b^{x_2}}{a\,b^{x_1}} \]

Simplify the right-hand side using the exponent rule \(b^{x_2}/b^{x_1} = b^{x_2-x_1}\):

\[ \frac{y_2}{y_1} = b^{\,x_2-x_1} \]

3. Now solve for \(b\). Take a positive real \((x_2-x_1)\)-th root (equivalently use logarithms). The algebraic solution is:

\[ b = \left(\frac{y_2}{y_1}\right)^{\! \frac{1}{x_2-x_1}}. \]

Explanation: raising both sides to the power \(1/(x_2-x_1)\) inverts the exponent on the right-hand side.

4. Having found \(b\), substitute back into one of the original equations to solve for \(a\). Using \((x_1,y_1)\):

\[ a = \frac{y_1}{b^{x_1}}. \]

5. Summary formula (from two points):

\[ b = \left(\frac{y_2}{y_1}\right)^{\! \frac{1}{x_2-x_1}}, \qquad a = \frac{y_1}{b^{x_1}}. \]

6. Note: This requires \(y_1\) and \(y_2\) have the same sign and \(y_1\neq 0\), so the ratio \(y_2/y_1\) is positive and a real root exists with \(b>0\).

3. Solving for x when a and b are known (finding the input that gives a certain output)

If you are given a particular output \(y\) and you know \(a\) and \(b\), solve for \(x\) by isolating the exponential and using logarithms. Steps:

1. Start with \[ y = a\,b^x. \]
2. Divide both sides by \(a\) (assuming \(a\neq 0\)):

\[ \frac{y}{a} = b^x. \]

3. Take logarithms. You can use natural logarithm ln or logarithm base 10 or any base; using ln is common:

\[ \ln\!\left(\frac{y}{a}\right) = \ln\!\big(b^x\big). \]

4. Use the log power rule \(\ln(b^x) = x \ln(b)\) to bring \(x\) down:

\[ \ln\!\left(\frac{y}{a}\right) = x\,\ln(b). \]

5. Solve for \(x\):

\[ x = \frac{\ln\!\left(\dfrac{y}{a}\right)}{\ln(b)}. \]

6. Remarks: this formula is valid when \(\dfrac{y}{a}>0\) and \(b>0,\, b\neq 1\).

4. Example (illustrative; every step shown)

Suppose you are given two points on an exponential curve: \((1,6)\) and \((4,48)\). Find \(a\) and \(b\).

1. Write equations:

\[ 6 = a\,b^1 = a\,b \]

\[ 48 = a\,b^4 \]

2. Divide the second by the first to eliminate \(a\):

\[ \frac{48}{6} = \frac{a\,b^4}{a\,b} \]

Simplify both sides:

\[ 8 = b^{4-1} = b^3 \]

3. Solve for \(b\) by taking cube root:

\[ b = \sqrt[3]{8} = 2 \]

4. Substitute \(b=2\) into \(6 = a\,b\) to find \(a\):

\[ 6 = a\times 2 \]

\[ a = \frac{6}{2} = 3 \]

5. Conclusion: the exponential function is

\[ y = 3\cdot 2^x. \]

5. Properties and useful transformations

1. Logarithmic form: taking logarithms of both sides gives \(\ln y = \ln a + x\ln b\). This is a linear equation in \(x\) if you view \(\ln y\) versus \(x\). That fact is useful for fitting data: plotting \(\ln y\) against \(x\) should produce a straight line with slope \(\ln b\) and intercept \(\ln a\).
2. Growth rate: for small increments \(\Delta x=1\), the factor changes by \(b\). For continuous growth models you often convert to the base \(e\) form \(y = A e^{kx}\) where \(A=a\) and \(b = e^{k}\), so \(k = \ln b\).
3. Behavior as \(x\to\infty\): if \(b>1\) then \(y\to\infty\) (growth). If \(0

6. Quick checklist when working with y = a b^x

• If you have the value at \(x=0\), set \(a\) equal to that value.
• If you have two points, divide the equations to find \(b\) and then back-substitute to find \(a\).
• To solve for \(x\), isolate \(b^x\) and take logarithms.
• Ensure \(b>0\) and that ratios used inside logarithms are positive.

This completes the detailed explanation and algebraic procedures for working with the exponential function \(y = a b^x\).