Let's address each part of the question:
1. \( f(x) = 3^x - 2 \)
This function is a transformation of the exponential function \( f(x) = 3^x \). It is shifted down by 2 units due to the "- 2" outside the exponent.
- Domain: The domain of exponential functions is always all real numbers.
- Range: Since the base is \( 3^x \), the range will be all positive real numbers shifted down by 2 units.
- Horizontal Asymptote: There's no horizontal asymptote in exponential functions.
- Y-intercept: To find the y-intercept, set \( x = 0 \) and solve for \( f(0) \).
\( f(0) = 3^0 - 2 = 1 - 2 = -1 \)
2. \( f(x) = 3^{-1} \)
This function is a constant function where \( f(x) \) is always equal to \( \frac{1}{3} \).
- Domain: All real numbers.
- Range: The range is just the constant value \( \frac{1}{3} \).
- Horizontal Asymptote: There's no horizontal asymptote in constant functions.
- Y-intercept: The y-intercept is where \( x = 0 \), so \( f(0) = \frac{1}{3} \).
Now, let's solve the equations:
3. \( 3 = 81 \)
This equation is not true. There is no solution since \( 3 \) does not equal \( 81 \).
4. \( 4 \cdot 0.2 \cdot 2^2 = 16^2 \)
Let's solve this:
\( 4 \cdot 0.2 \cdot 2^2 = 16^2 \)
\( 4 \cdot 0.2 \cdot 4 = 256 \)
\( 3.2 \cdot 4 = 256 \)
\( 12.8 = 256 \)
This equation is not true. There is no solution since \( 12.8 \) does not equal \( 256 \).
5. \( (e^1)^1 \cdot e^{22} = e^{12} \)
Let's solve this:
\( (e^1)^1 \cdot e^{22} = e^{12} \)
\( e \cdot e^{22} = e^{12} \)
\( e^{23} = e^{12} \)
Now, since the bases are the same, we equate the exponents:
\( 23 = 12 \)
This equation is not true. There is no solution since \( 23 \) does not equal \( 12 \).