Without using a calculator, find the exact value of
2 log3 6 − log3 4
3 marks
Show answer
2
2
Solve the equation 5x = 18, giving your answer to 3 significant figures.
3 marks
Show answer
x = 1.80
3
Find the exact solution of each equation.
e2x − 1 = 7
ln(3x + 2) = 4
4 marks
Show answer
a) x = (1 + ln 7)2b) x = (e4 - 2)3
4
Given that loga 2 = p and loga 3 = q, write each of the following in terms of p and q
loga 18
loga427
loga (12a)
5 marks
Show answer
a) p + 2qb) 2p - 3qc) 2p + q + 1
5
Solve log2(x + 5) − log2(x − 1) = 3, x > 1
4 marks
Show answer
x = 13/7
6
Show that the equation log3x + log3(x − 6) = 3 can be written as x2 − 6x − 27 = 0
Hence solve log3x + log3(x − 6) = 3, explaining why one root of the quadratic is not a solution.
5 marks
Show answer
a) log_3 x(x - 6) = 3, x(x - 6) = 27, x2 - 6x - 27 = 0b) x = 9 (x = -3 is rejected: log_3(-3) is undefined)
7
Sketch the graph of y = ex − 4
On your sketch show the coordinates of any points where the graph meets the coordinate axes, giving exact values, and state the equation of the asymptote.
4 marks
Show answer
through (0, -3) and (ln 4, 0); asymptote y = -4
8
Solve the equation 3x + 1 = 52x − 1
giving your answer in the form ln aln b, where a and b are rational numbers, and then to 3 significant figures.
5 marks
Show answer
x = ln 15 / ln (25/3) = 1.28
9
Show that the substitution u = 2x turns the equation 22x − 7(2x) + 12 = 0 into u2 − 7u + 12 = 0
Hence solve 22x − 7(2x) + 12 = 0, giving any non-exact answer to 3 significant figures.
6 marks
Show answer
a) 22x = (2x)2 = u2b) x = 2 or x = 1.58
10
Solve the simultaneous equations
log2x + log2y = 6
log2x − 2 log2y = 3
6 marks
Show answer
x = 32, y = 2
11
On the same axes, sketch the curves with equations y = e2x and y = 6 − ex, showing the coordinates of the points where each curve meets the y-axis and the equations of any asymptotes.
Find the exact coordinates of the point where the curves meet, explaining why there is only one such point.
7 marks
Show answer
a) y = e2x through (0, 1) with asymptote y = 0; y = 6 - ex through (0, 5) with asymptote y = 6b) (ln 2, 4); ex = -3 has no solution
12
Prove that log4x = 12 log2x for x > 0
Hence solve log2x + log4x = 6
Solve log2x = log4(x + 6), explaining why only one solution is valid.
9 marks
Show answer
a) x = 4y = 22y so log_2 x = 2y = 2 log_4 xb) x = 16c) x = 3 (x = -2 is rejected)
Every question here is original and written by Robinson Tuition. It follows the Edexcel AS specification and suits AQA and OCR AS maths too; this is not an exam-board paper.