Exponential Functions and Algebraic Equivalence

General Form of an Exponential Function

• An exponential function is written in the standard form:   $y = a b^x$

  • $a$ represents the initial value or constant multiplier, where $a \neq 0$.

  • $b$ represents the base or growth factor, where $b > 0$ and $b \neq 1$

  • $x$ represents the independent variable exponent.

  • $y$ represents the dependent variable output.

Problem 38: Exponential Function Passing Through (2, 2) and (4, 16)

• Given Information:

  • First point: $(x_1, y_1) = (2, 2)$

  • Second point: $(x_2, y_2) = (4, 16)$

• Step 1: Set up the system of equations using $y = a b^x$

  • Substituting the first point $(2, 2)$:     $2 = a b^2$

  • Substituting the second point $(4, 16)$:     $16 = a b^4$

• Step 2: Express $a$ in terms of $b$ for each equation

  • From the first equation:     $a = \frac{2}{b^2}$

  • From the second equation:     $a = \frac{16}{b^4}$

• Step 3: Equate the two expressions for $a$ and solve for $b$

  • Set the expressions equal to each other:     $\frac{2}{b^2} = \frac{16}{b^4}$

  • Cross-multiply to clear fractions:     $2 b^4 = 16 b^2$

  • Divide both sides by $2 b^2$ (assuming $b \neq 0$):     $b^2 = 8$

  • Take the principal square root of both sides (since base $b > 0$):     $b = \sqrt{8} = 2 \sqrt{2}$

• Step 4: Solve for $a$

  • Substitute $b^2 = 8$ back into $a = \frac{2}{b^2}$:     $a = \frac{2}{8} = \frac{1}{4}$

• Step 5: Write the final exponential equation

  • Substituting $a = \frac{1}{4}$ and $b = 2 \sqrt{2}$ into $y = a b^x$:     $y = \frac{1}{4} (2 \sqrt{2})^x$

  • Equivalent representation in terms of base $8$:     $y = \frac{1}{4} (8^{x/2})$

Problem 39: Exponential Function Passing Through (3, 3) and (6, 12)

• Given Information:

  • First point: $(x_1, y_1) = (3, 3)$

  • Second point: $(x_2, y_2) = (6, 12)$

• Step 1: Set up the system of equations using $y = a b^x$

  • Substituting the first point $(3, 3)$:     $3 = a b^3$

  • Substituting the second point $(6, 12)$:     $12 = a b^6$

• Step 2: Express $a$ in terms of $b$ for each equation

  • From the first equation:     $a = \frac{3}{b^3}$

  • From the second equation:     $a = \frac{12}{b^6}$

• Step 3: Equate the two expressions for $a$ and solve for $b$

  • Set the expressions equal to each other:     $\frac{3}{b^3} = \frac{12}{b^6}$

  • Cross-multiply to clear fractions:     $3 b^6 = 12 b^3$

  • Divide both sides by $3 b^3$:     $b^3 = 4$

  • Take the cube root of both sides:     $b = \sqrt[3]{4} = 4^{1/3}$

• Step 4: Solve for $a$

  • Substitute $b^3 = 4$ back into $a = \frac{3}{b^3}$:     $a = \frac{3}{4}$

• Step 5: Write the final exponential equation

  • Substituting $a = \frac{3}{4}$ and $b = \sqrt[3]{4}$ into $y = a b^x$:     $y = \frac{3}{4} (\sqrt[3]{4})^x$

  • Equivalent exponential representation:     $y = \frac{3}{4} (4^{x/3})$

Problem 40: Proving Equivalence of Two Algebraic Expressions

• Given Expressions:

  • First Expression:     $a = \frac{e^2}{e^3 + e^2}$

  • Second Expression:     $a = \frac{1}{1 + e^{3-2}}$

• Proof of Equivalence:

  • Begin with the first expression:     $a = \frac{e^2}{e^3 + e^2}$

  • Factor out $e^2$ from the denominator:     $e^3 + e^2 = e^2 (e + 1) = e^2 (e^{3-2} + 1)$

  • Divide both numerator and denominator by $e^2$:     $\frac{e^2 / e^2}{(e^3 + e^2) / e^2} = \frac{1}{\frac{e^3}{e^2} + \frac{e^2}{e^2}}$

  • Apply exponent rules for quotient of powers, where $\frac{e^m}{e^n} = e^{m-n}$:     $\frac{e^3}{e^2} = e^{3-2}$     $\frac{e^2}{e^2} = 1$

  • Substitute these simplified terms back into the fraction:     $a = \frac{1}{e^{3-2} + 1} = \frac{1}{1 + e^{3-2}}$

  • Simplifying further gives:     $a = \frac{1}{1 + e^1} = \frac{1}{1 + e}$

  • This demonstrates that both algebraic expressions are identical and equivalent.