Rewrite the log as a power question: _2 8 = ? \;\; 2^? = 8 Write 8 as a power of 2: 8 = 2^3 Read off the exponent: 2^3 = 8 \;\; _2 8 = 3 Answer: _2 8 = 3

Which statement correctly explains why _2 8 = 3?

Recall the meaning of a logarithm: _a b = x \;\; a^x = b Apply it to this case: 2^3 = 8 \;\; _2 8 = 3 Picture the inverse relationship: y = 2^x and y = _2 x are reflections in y=x Answer: _2 8 = 3 because 2^3 = 8

Which graph shows y = _2 x for x > 0?

Recall key features of y = log_2 x: x-intercept at (1,0),\; undefined for x≤ 0 Note the growth and asymptote: increasing; y- as x 0^+ Match to the correct sketch: y = _2 x Rule out the exponential graph: y = 2^x passes through (0,1), not (1,0) Rule out reflections: y = -_2 x is decreasing Confirm the key point: _2 1 = 0 (1,0) lies on the curve Answer: y = _2 x

Solve _10 x - 6_x 10 = 1.

Rewrite log_x 10 using change of base: _x 10 = 1_10 x Substitute a letter: Let u = _10 x Rewrite the equation: u - (6)/(u) = 1 Multiply through by u: u^2 - 6 = u Form a quadratic: u^2 - u - 6 = 0 Factorise: (u-3)(u+2) = 0 Solve for u: u = 3 u = -2 Undo the first: _10 x = 3 \;\; x = 10^3 = 1000 Undo the second: _10 x = -2 \;\; x = 10^-2 = 0.01 State the solutions: x = 1000 x = 0.01 Answer: x = 1000 or x = 0.01

Solve the simultaneous equations _2 x + _2 y = 5 and (_2 x)(_2 y) = 6.

Introduce two letters: Let a = _2 x,\; b = _2 y Translate the first equation: a + b = 5 Translate the second equation: ab = 6 Recognise sum and product of roots: t^2 - (a+b)t + ab = 0 Form the quadratic: t^2 - 5t + 6 = 0 Factorise: (t-2)(t-3) = 0 Find a and b: \a,b\ = \2,3\ Take the first case: a=2, b=3:\; _2 x = 2,\; _2 y = 3 Convert to x and y: x = 2^2 = 4, y = 2^3 = 8 Take the second case: a=3, b=2:\; x = 2^3 = 8,\; y = 2^2 = 4 Check the first pair: _2 4+_2 8=2+3=5; (2)(3)=6\; Check the second pair: _2 8+_2 4=3+2=5; (3)(2)=6\; Explain why both are valid: the equations are symmetric in x,y Note the key idea: sum and product quadratic in t State the final answer: (x,y) = (4,8) \;or\; (8,4) Answer: (x,y) = (4,8) or (8,4)

A-Level Logarithms Practice Questions

Free A-Level Logarithms practice questions with full step-by-step worked solutions. Covers logarithm as inverse, evaluate, base 10, log as inverse. Practise exam-style problems and check your method.

logarithm as inverseevaluatebase 10log as inverseconverting formnegative index

A-Level70 questionsStep-by-step solutions

Question 1

1 markeasy

Evaluate

\log_2 8

Show worked solution

Worked solution

Rewrite the log as a power question
$\log_2 8 = ? \;\Rightarrow\; 2^{?} = 8$
A logarithm just asks 'what power do I raise the base to?' Here the base is 2, so we ask what power of 2 gives 8. This is the inverse of raising 2 to a power.
Write 8 as a power of 2
$8 = 2^3$
We list powers of 2: 2, 4, 8. So 8 is the third one, meaning 8 = 2 cubed. Recognising powers of small numbers is a key skill from indices.
Read off the exponent
$2^3 = 8 \;\Rightarrow\; \log_2 8 = 3$
Since 2 to the power 3 equals 8, the logarithm is exactly that power, 3. So the answer is 3.

Answer

\log_2 8 = 3

Question 2

2 markseasy

Which statement correctly explains why

\log_2 8 = 3

Show worked solution

Worked solution

Recall the meaning of a logarithm
$\log_a b = x \;\Longleftrightarrow\; a^x = b$
A logarithm is the inverse of raising to a power. It answers 'what power of the base gives this number?'.
Apply it to this case
$2^3 = 8 \;\Rightarrow\; \log_2 8 = 3$
Because 2 raised to the power 3 equals 8, the log base 2 of 8 is exactly that power, 3.
Picture the inverse relationship
$y = 2^x \text{ and } y = \log_2 x \text{ are reflections in } y=x$
The exponential and the logarithm undo each other, so their graphs mirror across the line y = x. The point (3, 8) on y = 2^x matches (8, 3) on y = log_2 x.

Answer

\log_2 8 = 3 \text{ because } 2^3 = 8

Question 3

3 marksintermediate

Which graph shows

y = \log_2 x

for

x > 0

Show worked solution

Worked solution

Recall key features of y = log_2 x
$x\text{-intercept at } (1,0),\; \text{undefined for } x\le 0$
A log graph only exists for positive x, and it always passes through (1, 0) because log of 1 is 0.
Note the growth and asymptote
$\text{increasing; } y\to-\infty \text{ as } x\to 0^+$
As x gets close to 0 the curve plunges down towards the y-axis (a vertical asymptote), and it rises slowly for large x.
Match to the correct sketch
$y = \log_2 x$
The correct graph is increasing, passes through (1, 0), and hugs the y-axis on the left. It is the mirror image of y = 2^x in the line y = x.
Rule out the exponential graph
$y = 2^x \text{ passes through } (0,1),\text{ not } (1,0)$
A graph through (0, 1) that shoots steeply upwards is y = 2^x, the inverse function, so it is not the answer.
Rule out reflections
$y = -\log_2 x \text{ is decreasing}$
If the curve falls as x increases, it is a reflection such as y = -log_2 x, not the increasing log graph we want.
Confirm the key point
$\log_2 1 = 0 \Rightarrow (1,0) \text{ lies on the curve}$
The correct graph must cross the x-axis exactly at x = 1, because log of 1 is 0 for any base.

Answer

y = \log_2 x

Question 4

7 markshard

Solve

\log_{10} x - 6\log_x 10 = 1

Show worked solution

Worked solution

Rewrite log_x 10 using change of base
$\log_x 10 = \dfrac{1}{\log_{10} x}$
The reciprocal rule says log_x 10 is one over log_10 x. This puts everything in one base.
Substitute a letter
$\text{Let } u = \log_{10} x$
This makes the equation easier to read.
Rewrite the equation
$u - \dfrac{6}{u} = 1$
Replace log_10 x by u and log_x 10 by 1/u.
Multiply through by u
$u^2 - 6 = u$
Clearing the fraction removes the denominator (u is not 0).
Form a quadratic
$u^2 - u - 6 = 0$
Move everything to one side.
Factorise
$(u-3)(u+2) = 0$
Two numbers multiply to -6 and add to -1: they are -3 and +2.
Solve for u
$u = 3 \quad\text{or}\quad u = -2$
Both are allowed because log_10 x can be positive or negative.
Undo the first
$\log_{10} x = 3 \;\Rightarrow\; x = 10^3 = 1000$
Convert back to x.
Undo the second
$\log_{10} x = -2 \;\Rightarrow\; x = 10^{-2} = 0.01$
A negative log gives a value between 0 and 1.
State the solutions
$x = 1000 \quad\text{or}\quad x = 0.01$
Both are positive, so both are valid.

Answer

x = 1000 \text{ or } x = 0.01

Question 5

8 markschallenging

Solve the simultaneous equations

\log_2 x + \log_2 y = 5

and

(\log_2 x)(\log_2 y) = 6

Show worked solution

Worked solution

Introduce two letters
$\text{Let } a = \log_2 x,\; b = \log_2 y$
Working with a and b turns the log equations into simple number equations.
Translate the first equation
$a + b = 5$
log_2 x + log_2 y is a + b, which equals 5.
Translate the second equation
$ab = 6$
The product of the two logs is ab, which equals 6.
Recognise sum and product of roots
$t^2 - (a+b)t + ab = 0$
If we know the sum and product of two numbers, they are the roots of this quadratic. This idea comes from quadratics.
Form the quadratic
$t^2 - 5t + 6 = 0$
Substitute the sum 5 and product 6.
Factorise
$(t-2)(t-3) = 0$
Two numbers multiply to 6 and add to 5: they are 2 and 3.
Find a and b
$\{a,b\} = \{2,3\}$
So one log is 2 and the other is 3, in some order.
Take the first case
$a=2,\,b=3:\; \log_2 x = 2,\; \log_2 y = 3$
This assigns 2 to x and 3 to y.
Convert to x and y
$x = 2^2 = 4,\quad y = 2^3 = 8$
Convert each log back to a value.
Take the second case
$a=3,\,b=2:\; x = 2^3 = 8,\; y = 2^2 = 4$
Swapping the roles gives the second solution pair.
Check the first pair
$\log_2 4+\log_2 8=2+3=5;\quad (2)(3)=6\;\checkmark$
For (4, 8): the logs are 2 and 3, which sum to 5 and multiply to 6, matching both equations.
Check the second pair
$\log_2 8+\log_2 4=3+2=5;\quad (3)(2)=6\;\checkmark$
For (8, 4): the logs are 3 and 2, which again sum to 5 and multiply to 6.
Explain why both are valid
$\text{the equations are symmetric in } x,y$
Swapping x and y leaves both equations unchanged, so if one pair works then the swapped pair works too.
Note the key idea
$\text{sum and product} \to \text{quadratic in } t$
Knowing the sum and product of two numbers let us build a quadratic whose roots are those numbers, a neat link back to quadratics.
State the final answer
$(x,y) = (4,8) \;\text{or}\; (8,4)$
Both ordered pairs are solutions of the simultaneous equations.

Answer

(x,y) = (4,8) \text{ or } (8,4)

Unlock 65 more Logarithms questions

Create a free account to work through every A-Level Logarithms question with instant step-by-step worked solutions, progress tracking and interactive lessons.

Full worked solutions for every question
Interactive lessons and instant feedback
Track your mastery across every topic

Create a Free Account

No card required · Free forever

More Logarithms practice

Hard Questions Worked Solutions

A-Level Logarithms Practice Questions

Worked solution

Worked solution

Worked solution

Worked solution

Worked solution

Unlock 65 more Logarithms questions

More Logarithms practice

Related Pure Maths topics