Exponential Inequalities, Equivalent Exponential Functions, and Graphing Transformations

Given Inequality: $9^x < 4^x$
Graphical Analysis:
- Both $y_1 = 9^x$ and $y_2 = 4^x$ are strictly increasing exponential functions that intersect at $(0, 1)$ .
- For $x > 0$ , $9^x$ grows faster than $4^x$ , so $9^x > 4^x$
- For $x < 0$ , $9^x$ decays to $0$ faster than $4^x$ , meaning $9^x < 4^x$
Algebraic Derivation:
- Divide both sides by $4^x$ (which is strictly positive for all real $x$ ): $\left(\frac{9}{4}\right)^x < 1$
- Take the natural logarithm of both sides: $\ln\left(\left(\frac{9}{4}\right)^x\right) < \ln(1)$ $x \cdot \ln\left(\frac{9}{4}\right) < 0$
- Since $\frac{9}{4} > 1$ , $\ln\left(\frac{9}{4}\right) > 0$ . Dividing by a positive value yields: $x < 0$
Solution Set:
- Interval Notation: $(-\infty, 0)$
- Key Point of Intersection: $(0, 1)$

Given Inequality: $6^{-x} > 8^{-x}$
Rewriting with Positive Bases: $\left(\frac{1}{6}\right)^x > \left(\frac{1}{8}\right)^x$
Graphical Analysis:
- Both functions decay toward $0$ as $x \rightarrow \infty$ and intersect at $(0, 1)$ .
- Since $\frac{1}{6} > \frac{1}{8}$ , raising $\frac{1}{6}$ to a positive exponent $x > 0$ yields a value greater than raising $\frac{1}{8}$ to that same positive exponent.
- For negative values $x < 0$ , let $x = -k$ where $k > 0$ : $6^k > 8^k$ This is false for all $k > 0$ .
Solution Set:
- Interval Notation: $(0, \infty)$

Given Inequality: $\left(\frac{1}{4}\right)^x > \left(\frac{1}{3}\right)^x$
Algebraic Analysis:
- Expressing with base integers using negative exponents: $4^{-x} > 3^{-x}$
- Testing values:
- At $x = -1$ : $\left(\frac{1}{4}\right)^{-1} = 4$ and $\left(\frac{1}{3}\right)^{-1} = 3$ . Since $4 > 3$ , the inequality holds for negative values.
- At $x = 0$ : $1 > 1$ is false.
- At $x = 1$ : $\frac{1}{4} > \frac{1}{3}$ is false.
Solution Set:
- Interval Notation: $(-\infty, 0)$

Objective: Demonstrate using exponent rules that $y_1 = 3^{2x+4}$ and $y_2 = 9^{x+2}$ represent identical functions.
Relevant Exponent Property:
- Power of a Power Rule: $(a^m)^n = a^{m \cdot n}$
- Base Substitution: $9 = 3^2$
Proof:
- Begin with $y_2$ : $y_2 = 9^{x+2}$
- Substitute $3^2$ for $9$ : $y_2 = (3^2)^{x+2}$
- Apply the Power of a Power rule by multiplying exponents: $y_2 = 3^{2 \cdot (x+2)}$
- Distribute the exponent $2$ through the expression $(x+2)$ : $y_2 = 3^{2x+4}$
- Since $y_2 = 3^{2x+4} = y_1$ , the two functions are identical.

Objective: Demonstrate using exponent rules that $y_1 = 2(2^{3x-2})$ and $y_2 = 2^{3x-1}$ represent identical functions.
Relevant Exponent Property:
- Product Rule of Exponents: $a^m \cdot a^n = a^{m+n}$
- Explicit Exponent Rule: $2 = 2^1$
Proof:
- Begin with $y_1$ : $y_1 = 2(2^{3x-2})$
- Rewrite the leading coefficient $2$ as $2^1$ : $y_1 = 2^1 \cdot 2^{3x-2}$
- Apply the Product Rule of Exponents by adding exponents: $y_1 = 2^{1 + (3x-2)}$
- Combine constant terms in the exponent: $y_1 = 2^{3x - 2 + 1}$ $y_1 = 2^{3x-1}$
- Since $y_1 = 2^{3x-1} = y_2$ , the two functions are identical.

Constant Definition:
- Euler's number $e$ is an irrational constant: $e \approx 2.71828$
Key Features of $y = e^x$ :
- Domain: $(-\infty, \infty)$
- Range: $(0, \infty)$
- Horizontal Asymptote: $y = 0$
- Y-intercept: $(0, 1)$
Table of Key Points:
- At $x = -1$ : $y = e^{-1} = \frac{1}{e} \approx 0.368$
- At $x = 0$ : $y = e^0 = 1$
- At $x = 1$ : $y = e^1 = e \approx 2.71828$
- At $x = 2$ : $y = e^2 \approx 7.389$

Transformation Sequence from Parent Function $y = e^x$ :
1. Horizontal Shift: Shift right by $3$ units due to $(x - 3)$ .
2. Vertical Stretch: Stretch vertically by a factor of $2$ due to coefficient $2$
3. Vertical Shift: Shift up by $1$ unit due to constant term $+1$
Asymptote:
- The horizontal asymptote shifts vertically from $y = 0$ to $y = 1$
Domain and Range:
- Domain: $(-\infty, \infty)$
- Range: $(1, \infty)$
Calculated Key Reference Points:
- At $x = 3$ : $y = 2e^{3-3} + 1$ $y = 2e^0 + 1$ $y = 2(1) + 1 = 3$ Point: $(3, 3)$
- At $x = 4$ : $y = 2e^{4-3} + 1$ $y = 2e^1 + 1$ $y = 2e + 1 \approx 2(2.718) + 1 = 6.436$ Point: $(4, 2e + 1) \approx (4, 6.436)$
- At $x = 2$ : $y = 2e^{2-3} + 1$ $y = 2e^{-1} + 1$ $y = \frac{2}{e} + 1 \approx \frac{2}{2.718} + 1 \approx 1.736$ Point: $\left(2, \frac{2}{e} + 1\right) \approx (2, 1.736)$