The diagram shows three graphs A is part of the line y = x . B is part of the graph of f x = 3 2 x - 1 + 2 . C is the reflection of f x = 3 2 x - 1 + 2 in the line y = x i.e., g x = log 3 2 x - 2 + 1 . Write down the equation of vertical asymptote to the graph of y = g x .

Given function is g x = log 3 2 x - 2 + 1 The vertical asymptote occurs when the function is not defined. The vertical asymptote of the function log x occurs when x = 0 . Similarly g x is not defined when x - 2 = 0 ⇒ x = 2 Therefore, vertical asymptote of g x is x = 2 .

The diagram shows three graphs A is part of the line y = x . B is part of the graph of f x = 3 2 x - 1 + 2 . C is the reflection of f x = 3 2 x - 1 + 2 in the line y = x i.e., g x = log 3 2 x - 2 + 1 . Write down the equation of vertical asymptote to the graph of y = g x .

Given function is g x = log 3 2 x - 2 + 1 The vertical asymptote occurs when the function is not defined. The vertical asymptote of the function log x occurs when x = 0 . Similarly g x is not defined when x - 2 = 0 ⇒ x = 2 Therefore, vertical asymptote of g x is x = 2 .

The diagram shows three graphs A is part of the line y = x . B is part of the graph of f x = 3 2 x - 1 + 2 . C is the reflection of f x = 3 2 x - 1 + 2 in the line y = x i.e., g x = log 3 2 x - 2 + 1 . Hence state the domain of g .

Given function is g x = log 3 2 x - 2 + 1 The domain of a function g x is the set of all values for which the function is defined. We know that log x is defined when x > 0 Similarly g x is defined when x - 2 > 0 ⇒ x > 2 Therefore, domain of g x is x > 2 .

The diagram shows three graphs A is part of the line y = x . B is part of the graph of f x = 3 2 x - 1 + 2 . C is the reflection of f x = 3 2 x - 1 + 2 in the line y = x i.e., g x = log 3 2 x - 2 + 1 . Solve the equation f x + g x = 0 .

Given functions are f ( x ) = 3 2 x - 1 + 2 and g x = log 3 2 x - 2 + 1 g x can also be written as g x = log x - 2 log 3 2 + 1 Now, f ( x ) = - g ( x ) ⇒ 3 2 x - 1 + 2 = - log x - 2 log 3 2 + 1 So, graphically we can say that the value of x = 2 . 155 .

E M B I B E

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 9 - Representing Equivalent Quantities: Exponentials and Logarithms>Chapter review>Q 29 P2:

EASY

Diploma

IMPORTANT

Earn 100

The diagram shows three graphs

$A$ is part of the line $y = x$ . $B$ is part of the graph of $f (x) = {(\frac{3}{2})}^{x - 1} + 2$ . $C$ is the reflection of $f (x) = {(\frac{3}{2})}^{x - 1} + 2$ in the line $y = x$ i.e., $g (x) = \log_{\frac{3}{2}} (x - 2) + 1$ . Write down the equation of vertical asymptote to the graph of $y = g (x)$ .

Important Questions on Representing Equivalent Quantities: Exponentials and Logarithms

EASY

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 9 - Representing Equivalent Quantities: Exponentials and Logarithms>Chapter review>Q 29 P2:

The diagram shows three graphs

Question Image

$A$ is part of the line $y = x$ . $B$ is part of the graph of $f (x) = {(\frac{3}{2})}^{x - 1} + 2$ . $C$ is the reflection of $f (x) = {(\frac{3}{2})}^{x - 1} + 2$ in the line $y = x$ i.e., $g (x) = \log_{\frac{3}{2}} (x - 2) + 1$ . Hence state the domain of $g$ .

HARD

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 9 - Representing Equivalent Quantities: Exponentials and Logarithms>Chapter review>Q 29 P2:

The diagram shows three graphs

Question Image

$A$ is part of the line $y = x$ . $B$ is part of the graph of $f (x) = {(\frac{3}{2})}^{x - 1} + 2$ . $C$ is the reflection of $f (x) = {(\frac{3}{2})}^{x - 1} + 2$ in the line $y = x$ i.e., $g (x) = \log_{\frac{3}{2}} (x - 2) + 1$ . Solve the equation $f (x) + g (x) = 0$ .