Question

question 14 (1 point) saved
the function $f(x) = 4^x$ has a y-intercept of
\bigcirc a) 5
\bigcirc b) -1
\bigcirc c) 1
\bigcirc d) -5

question 15 (1 point)
solve for x: $\ln (x - 2) = 3$
\bigcirc a) $x = 2 + e^3$
\bigcirc b) x=5

Explanation:

Question 14 (assuming the function is \( f(x) = a^x \) or similar, let's correct and solve)

Step1: Recall y - intercept definition

The y - intercept of a function \( y = f(x) \) is the value of \( y \) when \( x = 0 \). So we need to find \( f(0) \).

Step2: Substitute \( x = 0 \) into the function

If the function is \( f(x)=a^{x} \) (assuming the missing base is non - zero, for example, if it's \( f(x) = 1^{x} \), \( f(0)=1^{0}=1 \); if it's \( f(x)=b^{x} \), in general, for any non - zero base \( b \), \( b^{0}=1 \)). Wait, maybe the original function was \( f(x)= - 1^{x} \) (but exponentiation order: \( - 1^{x}=-(1^{x}) \)). If \( x = 0 \), \( - 1^{0}=- 1 \). Let's assume the function is \( f(x)=-1^{x} \). Then \( f(0)=-1^{0}=-1 \). So the y - intercept is - 1, which is option b.

Step1: Recall the property of natural logarithm

The equation is \( \ln(x - 2)=3 \). We know that if \( \ln(a)=b \), then \( a = e^{b} \) (by the definition of the natural logarithm, where the base of \( \ln \) is \( e \)).

Step2: Apply the property to solve for \( x \)

Using the property \( \ln(a)=b\Rightarrow a = e^{b} \), for \( \ln(x - 2)=3 \), we have \( x - 2=e^{3} \).

Step3: Solve for \( x \)

Add 2 to both sides of the equation \( x - 2=e^{3} \), we get \( x=2 + e^{3} \).

Answer:

b) - 1

Question 15