Solving Exponential Equations

Question 1: Domain of a Composite Function (from file image_99a1b3.png)

The Question: If the domain of $f(x)$ is $(-3, 1)$ then the domain of $f(\ln x)$ is __________.

• $(e^{-1}, e^3)$
• $(0, \infty)$
• $(1, \infty)$
• $(e^{-3}, e^1)$

Detailed Solution:

We are finding the domain of the composite function $f(g(x))$ where the outer function is $f$ and the inner function is $g(x)=\ln x$.

1. Condition 1: $x$ must be in the domain of the inner function, $\ln x$.

   • The natural logarithm, $\ln x$, is only defined for positive numbers.
   • So, we must have $x > 0$.

2. Condition 2: The output of the inner function, $\ln x$, must be in the domain of the outer function, $f$.

   • The domain of $f$ is given as $(-3, 1)$. This means the input to $f$ must be greater than -3 and less than 1.
   • In our case, the input to $f$ is $\ln x$. So we have the inequality: $$-3 < \ln x < 1$$
   • We can split this into two parts and solve for $x$ by converting to exponential form (using base $e$):
     • $\ln x > -3 \implies x > e^{-3}$
     • $\ln x < 1 \implies x < e^1$

3. Combine all conditions:

   • We need to find the intersection of $x > 0$, $x > e^{-3}$, and $x < e^1$.
   • Since $e^{-3} = 1/e^3$ is a small positive number, the condition $x > e^{-3}$ is more restrictive than $x > 0$.
   • So our final combined condition is $e^{-3} < x < e^1$.

Final Answer: The domain is $(e^{-3}, e^1)$.

Question 2: Analyzing $f(x) = \ln(e^x - e^{-x})$ (from file image_99a1b3.png)

The Question: Consider the function $f(x) = \ln(e^x - e^{-x})$. Find the correct options. (Multiple Select Question)

Detailed Solution:

Let’s analyze the properties of this function.

• Domain: The argument of the natural logarithm must be strictly positive.

  • $e^x - e^{-x} > 0$
  • $e^x > e^{-x}$
  • Since $y=e^x$ is an increasing function, we can compare the exponents directly:
  • $x > -x$
  • $2x > 0 \implies x > 0$.
  • The domain is $(0, \infty)$. Therefore, the statement “The domain of $f$ is $\mathbb{R}$” is FALSE.

• Is the function increasing? We can find the derivative to check the slope. Using the chain rule, the derivative of $\ln(u)$ is $u’/u$.

  • Let $u = e^x - e^{-x}$. Then $u’ = e^x - (-e^{-x}) = e^x + e^{-x}$.
  • $f’(x) = \frac{u’}{u} = \frac{e^x + e^{-x}}{e^x - e^{-x}}$.
  • For any $x$ in the domain ($x>0$), the numerator ($e^x + e^{-x}$) is always positive. The denominator ($e^x - e^{-x}$) is also positive.
  • Since the derivative $f’(x)$ is always positive on the domain, the function is an increasing function.

• Is the function one-to-one?

  • A function that is strictly increasing over its entire domain will always pass the Horizontal Line Test.
  • Since we’ve shown the function is always increasing, it is a one-to-one function.

Final Answer: The correct options are:

• $f$ is an one-one function
• $f$ is an increasing function

Question 3: Solving an Exponential Equation (from file image_99a1b3.png)

The Question: Solve for $x$ in $3^{(x+2)} = 243$.

Core Concept: If you can express both sides of an equation as a power of the same base ($b^A = b^B$), you can equate the exponents ($A=B$).

Detailed Solution:

1. The equation is $3^{x+2} = 243$. The base on the left is 3. We need to write 243 as a power of 3.
2. Calculate powers of 3:
   • $3^1 = 3$
   • $3^2 = 9$
   • $3^3 = 27$
   • $3^4 = 81$
   • $3^5 = 243$
3. Rewrite the equation with the same base:
   • $3^{x+2} = 3^5$
4. Equate the exponents:
   • $x + 2 = 5$
5. Solve for x:
   • $x = 5 - 2 = 3$

Final Answer: The value of $x$ is 3.

Question 4: Solving a Complex Exponential Equation (from file image_99a153.png)

The Question: Solve for $x$ in $e^{(4x^2 - 2x - 12)} = 1$. (Note: Based on the options and standard problem types, the exponent is interpreted as a quadratic $4x^2 - 2x - 12$.)

Core Concept: The only way for $e$ raised to a power to equal 1 is if that power is 0, because $e^0 = 1$.

Detailed Solution:

1. Set the exponent equal to zero:

   • Given the equation $e^{(4x^2 - 2x - 12)} = 1$, we can immediately conclude:
   • $4x^2 - 2x - 12 = 0$

2. Simplify the quadratic equation:

   • Divide the entire equation by 2 to make it easier to work with:
   • $2x^2 - x - 6 = 0$

3. Solve the quadratic equation. We can use factoring.

   • We are looking for two numbers that multiply to $a \cdot c = 2 \cdot (-6) = -12$ and add up to $b = -1$.
   • The numbers are -4 and +3.
   • Rewrite the middle term: $2x^2 - 4x + 3x - 6 = 0$.
   • Factor by grouping: $(2x^2 - 4x) + (3x - 6) = 0$.
   • $2x(x - 2) + 3(x - 2) = 0$.
   • $(2x + 3)(x - 2) = 0$.

4. Find the possible values for x:

   • Either $2x + 3 = 0 \implies x = -3/2$
   • Or $x - 2 = 0 \implies x = 2$

5. Check the options: The options are all integers, so we select the integer solution.

Final Answer: The value of $x$ is 2.