Logarithmic & Exponential Functions (Cambridge International IGCSE Further Maths): Exam Questions

Given that $\mathrm{log}_{a}p+\mathrm{log}_{a}5--\mathrm{log}_{a}4=\mathrm{log}_{a}20$, find the value of$p$.

In this question, $a,b,c$ and $d$ are positive constants.

(i) It is given that $y = log_{a} ( x + 3 ) + log_{a} ( 2 x - 1 )$. Explain why $x$ must be greater than $\frac{1}{2}$.

(ii) Find the exact solution of the equation $\frac{log_{a} 6}{log_{a} ( y + 3 )} = 2$

The function$f$ is defined by$f(x)=\mathrm{ln}(2x+1)\mathrm{for}x\geq0$.

Sketch the graph of $y=f(x)$ and hence sketch the graph of $y=f^{-1}(x)$ on the axes below.

Solve the equation $\frac{9^{5x}}{27^{x-2}}=243$

Write $3\mathrm{lg}x+2-\mathrm{lg}y$ as a single logarithm.

Using the substitution $y=2^{x}$ , or otherwise, solve $2^{2x+1}-2^{x+1}-2^{x}+1=0$.

Solve the following equation.

$\frac{1+3e^{x}}{e^{x}+e^{-x}}=1$

Give your answer(s) in the form $p\mathrm{ln}q$ where $p$ and $q$ are integers.

Given that $\mathrm{log}_{2}y=c$, express $log_{2} ( 2 y ) - log_{y} 2$ in terms of $c$ only.

For variables $x$ and $y$, plotting $\mathrm{ln}y$ against $\mathrm{ln}x$ gives a straight-line graph passing through the points $(6,5)$ and $(8,9)$. Show that $y$ = $e^{p}x^{q}$ where $p$ and $q$ are integers to be found.

Variables $x$ and $y$ are such that, when $\mathrm{lg}y$ is plotted against $x^{3}$, a straight line graph passing through the points (6,7) and (10,9) is obtained. Find $y$ as a function of $x$.

Given that $\mathrm{log}_{2}x+2\mathrm{log}_{4}y=8$, find the value of $xy$.

Variables $x$ and $y$ are connected by the relationship $y=Ax^{n}$, where $A$ and $n$ are constants.

Transform the relationship $y=Ax^{n}$ to straight line form.

Extract:

Variables x and y are connected by the relationship y = Ax^n, where A and n are constants.

Transform the relationship y = Ax^n to straight line form.

Take the natural logarithm of both sides. ln y = ln(Ax^n)

Use logarithm laws to rewrite. ln y = ln A + ln x^n

Use the logarithm power rule to rewrite in the form Y = mX + c ln y = n ln x + ln A

When $\mathrm{ln}y$ is plotted against $\mathrm{ln}x$ a straight line graph passing through the points (0, 0.5) and (3.2, 1.7) is obtained.

Find the value of $n$ and of $A$.

Extract:

Variables x and y are connected by the relationship y = Ax^n, where A and n are constants.

When ln y is plotted against ln x a straight line graph passing through the points (0, 0.5) and (3.2, 1.7) is obtained.

From the mark scheme for part b: ln A = 0.5 A = e^0.5 n = (1.7 - 0.5)/(3.2 - 0) = 3/8

Find the value of $y$ when $x=11$.

The population $P$, in millions, of a country is given by $P=A\times b^{t}$, where $t$ is the number of years after January 2000 and $A$ and $b$ are constants. In January 2010 the population was 40 million and had increased to 45 million by January 2013.

Show that $b=$1.04 to 2 decimal places and find $A$ to the nearest integer.

Extract:

The population P, in millions, of a country is given by P = A × b^t, where t is the number of years after January 2000 and A and b are constants. In January 2010 the population was 40 million and had increased to 45 million by January 2013.

b = 1.04 to 2 decimal places and A = 27 to the nearest integer.

Find the population in January 2020, giving your answer to the nearest million.

Extract:

The population P, in millions, of a country is given by P = A × b^t, where t is the number of years after January 2000 and A and b are constants. In January 2010 the population was 40 million and had increased to 45 million by January 2013.

From the solution: b = 1.04 and A = 27

In January of which year will the population be over 100 million for the first time?

The number, $b$, of bacteria in a sample is given by $b=P+Qe^{2t}$ , where $P$ and $Q$ are constants and $t$ is time in weeks. Initially there are 500 bacteria which increase to 600 after 1 week.

Find the value of $P$ and of $Q$.

Extract:

The number, b, of bacteria in a sample is given by b = P + Q*e^(2t), where P and Q are constants and t is time in weeks. Initially there are 500 bacteria which increase to 600 after 1 week.

Find the value of P and of Q.

Find the number of bacteria present after 2 weeks.

Extract:

The number, b, of bacteria in a sample is given by b = P + Qe^(2t), where P and Q are constants and t is time in weeks. Initially there are 500 bacteria which increase to 600 after 1 week.

From the mark scheme: Q = 100/(e² - 1) = 15.651... = 15.7 (3 sf) P = 500 - 15.65... = 484.34... = 484 (3 sf)

Find the first week in which the number of bacteria is greater than 1 000 000.

$f(x)=3+e^{x}\mathrm{for}x\inℝ$ $g(x)=9x-5\mathrm{for}x\inℝ$

Find the exact solution of $f^{-1}(x)=g'(x)$ .

A mathematical model is used to predict the population of an island, $P$, over time, $t$ years.

A plot of $\mathrm{ln}P$ against $\mathrm{ln}t$ gives a straight line that is the perpendicular bisector of the points $( 0 , 11 )$ and $( 6 , 7 )$, as shown.

Find and simplify a formula for $P$ in terms of $t$.

Solve the equation $3^{2x+1}+8(3^{x})--3=0$.

Solve the equation$4\mathrm{log}_{y}2+\mathrm{log}_{2}y=4$ .

Write the expression $\mathrm{log}_{a}9+(\mathrm{log}_{a}b)(\mathrm{log}_{\sqrt{b}}9a)$ in the form $c+d\mathrm{log}_{a}9$, where $c$ and $d$ are integers.

Find the exact solution of $3^{2x}-3^{x+1}-4=0$.

Solve the simultaneous equations

$10^{x+2y}=5$,

$10^{3x+4y}=50$ ,

giving $x$ and $y$ in exact simplified form.

$\mathrm{log}_{a}\sqrt{b}-\frac{1}{2}=\mathrm{log}_{b}a,\mathrm{where}a>0\mathrm{and}b>0$.

Solve this equation for $b$, giving your answers in terms of $a$.

Solve the simultaneous equations.

$log_{3} ( x + y ) = 2$

$2 log_{3} ( x + 1 ) = log_{3} ( y + 2 )$

It is known that $y=A\times10^{bx^{2}}$, where $A$ and $b$ are constants. When $\mathrm{lg}y$ is plotted against $x^{2}$, a straight line passing through the points (3.63, 5.25) and (4.83, 6.88) is obtained.

Find the value of $A$ and of $b$.

Extract:

It is known that y = A x 10 ^ b x ^2 , where A and b are constants. When lg y is plotted against x ^2, a line passing through the points (3.63, 5.25) and (4.83, 6.88) is obtained.

From the mark scheme: A equals 2.09 b equals 1.36

Using your values of $A$ and $b$, find the value of $y$ when $x=2$,

Extract:

It is known that y = A × 10^(bx²), where A and b are constants.

A = 2.09 b = 1.36

Using these values of A and b: y = 2.09 × 10^(1.36x²)

Find the positive value of $x$ when $y=4$.

$log_{2} ( y + 1 ) = 3 - 2 log_{2} x$

$log_{2} ( x + 2 ) = 2 + log_{2} y$

Show that $x^{3}+6x^{2}-32=0$.

Find the roots of $x^{3}+6x^{2}-32=0$.

Extract:

log₂(y+1) = 3 - 2log₂x

log₂(x+2) = 2 + log₂y

The roots to the equation x³ + 6x² - 32 = 0 are:

x = 2, x = -4, x = -4

Give a reason why only one root is a valid solution of the logarithmic equations. Find the value of $y$ corresponding to this root.