Solve a logarithmic equation

In 6 of 20 recent exams
Paper 1, no calculator
2 to 5 marks

How it is set in the exam

Paper 1 (no calculator) only, 2–5 marks per part: a whole Section A question, or the zeros part of a Section B function question, often straight after a "show that". Common (6 of 20 exam versions, 8 parts, all Paper 1 (no calculator)). It is also the last step inside calculus and sequence questions (Where exponents and logs hide).

Spot it in the exam

Exam phrases to spot (6)

“solve $\log_b(\ldots) - \log_b(\ldots) = c$ ”
“find the zeros of $f$ ”
“find the $x$ -intercepts”
“solve the simultaneous equations”
“hence solve”
“show that”

In disguise (3)

the zeros of the log of a quadratic: the argument equals $1$
"show that $y = \ldots$ ", a relation between two unknowns from a log equation with a constant term: the first part of a simultaneous pair
two bases that are powers of the same number, such as $\log_2$ and $\log_8$

Variants (5)

a single log $= 0$ : the $x$ -intercept of $\ln$ of a quadratic; the zeros of $\log_k$ of a quadratic, with surd answers
the difference of two logs in one base $=$ a constant, after a factorise-and-cancel "show that"
$\log A = \log B$ with a rational power, after a "show that"
two bases that are powers of one number, one term with a coefficient
simultaneous log equations, one with the unknown as the base, with a relation shown first

Looks like it, but isn't (7)

a quadratic in $\ln x$ (terms in $(\ln x)^2$ and $\ln x$ ): Quadratic in disguise
the domain of the log of a quadratic (argument $> 0$ ): Quadratic inequalities; of the log of a linear expression: Exponential and log graphs
the greatest value of the log of a quadratic: Find the vertex
the unknown only in a power: Solve an exponential equation
the vertical asymptote of a log graph: Exponential and log graphs
an equation that comes from an integral with an unknown limit: Integrate: choose the method (its last steps are this type's)
a log-scale model solved for its input: Exponential and log models

Worked example

The zero of ln of a quadratic: surds, then the domain

Paper 1 (no calculator)

Question

Let $f(x) = \ln\left(x^2 + 2x - 3\right)$ , for $x > 1$ . Find the coordinates of the point where the graph of $f$ meets the $x$ -axis. [4]

Try it yourself first, then compare.

Now try one like it

Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

Try one like it

Which method?

After one base and one log on each side:

$\log_a X = c$
Remove the logs by
$X = a^c$
Example
$\log_2(x + 3) = 5 \Rightarrow x + 3 = 32$
$\log_a X = \log_a Y$
Remove the logs by
$X = Y$
Example
$\ln(2x) = \ln(x + 4) \Rightarrow 2x = x + 4$
$\log_x c = d$ (unknown base)
Remove the logs by
$x^d = c$ , a power equation (the unknown-base row of Solve an exponential equation)
Example
$\log_x 32 = \frac{5}{2} \Rightarrow x^{\frac{5}{2}} = 32$ , so $x = 4$
terms in $(\ln x)^2$
Remove the logs by
Quadratic in disguise
Example
$(\ln x)^2 - 3\ln x + 2 = 0$

Form	Remove the logs by	Example
$\log_a X = c$	$X = a^c$	$\log_2(x + 3) = 5 \Rightarrow x + 3 = 32$
$\log_a X = \log_a Y$	$X = Y$	$\ln(2x) = \ln(x + 4) \Rightarrow 2x = x + 4$
$\log_x c = d$ (unknown base)	$x^d = c$ , a power equation (the unknown-base row of Solve an exponential equation)	$\log_x 32 = \frac{5}{2} \Rightarrow x^{\frac{5}{2}} = 32$ , so $x = 4$
terms in $(\ln x)^2$	Quadratic in disguise	$(\ln x)^2 - 3\ln x + 2 = 0$

In every case, test the roots in the original arguments, and against any domain the question states.

Steps

One base.
Change of base where the bases differ: $\log_8 x = \frac{1}{3}\log_2 x$ .
One log on each side
(or one log equal to a constant): power law first, then product and quotient. Write a constant as a log if that helps: $1 = \log_2 2$ .
Remove the logs
(table above).
Solve
the equation that is left: linear, quadratic, or a rational power.
Two equations:
remove the logs from each (a relation with no logs from the known base; a power equation from an unknown base), then substitute one into the other (Example 4.4).
Test every root in every original argument
(each must be $> 0$ ) and against any domain the question states (in Example 4.1 both roots make the argument $1$ , so only the domain can reject one); a base must also be positive and not $1$ . Reject the roots that fail, and say why. Use the equation as it was given: the laws can change which roots are allowed.

Common mistakes

Testing a root in a rewritten equation instead of the original.
$2\ln x = \ln 9$ has only $x = 3$ , because $\ln(-3)$ does not exist. But $\ln\left(x^2\right) = \ln 9$ has $x = 3$ and $x = -3$ , because $(-3)^2 = 9 > 0$ . After the power law the two equations look the same; their domains are not. (Compare $k^2 = 4$ in a binomial question, where both signs stay unless a condition removes one: Find an unknown (a constant or n) from given coefficients.)
Keeping a rejected root
writing $x = -1 - \sqrt{5}$ next to the right answer in Example 4.1 scores .
The $x$ -value alone when a point is asked
"the point where the graph meets the $x$ -axis" needs $\left(\sqrt{5} - 1, 0\right)$ . Markschemes have given for an $x$ -value on its own when a point was asked for; in Example 4.1, writing only $x = \sqrt{5} - 1$ would be that mistake.

More mistakes (8)

Rejecting a root just because it is negative
in Example 4.2, $x = -4$ is valid. What decides is the sign of each original argument and any domain the question states (in Example 4.1, $x > 1$ ), never the sign of $x$ on its own.
Removing the logs term by term
$\log_4 X - \log_4 Y = 1$ does not give $X - Y = 4$ . Combine into one log first.
Losing the factor from a change of base
$\log_8 x = \frac{1}{3}\log_2 x$ , so $6\log_8 x = 2\log_2 x$ , not $6\log_2 x$ and not $\log_2 x^6$ . Then the power law comes before the quotient law: $\log_2(3x) - 2\log_2 x = \log_2 \frac{3x}{x^2}$ , not $\log_2 \frac{3x}{2x}$ .
Removing the logs with a constant still outside
$\log_2 y = \log_2 x^2 - 3$ does not give $y = x^2 - 3$ . Write $3 = \log_2 8$ and combine first: $y = \frac{x^2}{8}$ (Example 4.4).
Checking instead of showing
in Example 4.4(a), substituting $y = \frac{x^2}{8}$ back into $\log_2 y = 2\log_2 x - 3$ is a check, not a proof. Start from the given equation and end on $x^2 = 8y$ (the LHS-to-RHS layout).
Reading $\ln X = 0$ as $X = 0$
it means $X = e^0 = 1$ .
The wrong base in the exponential form
$\log_2 X = 3$ means $X = 2^3$ , not $3^2$ and not $e^3$ .
An unknown base of $1$ or $0$
$\log_1$ and $\log_0$ don't exist, so such a root is rejected when the unknown is a base.

Easy to confuse with

Next typeExact value of a logarithm

Now try one like it

Keep going

Easy to confuse with