MHF4U Solving Exponential and Logarithmic Equations: Notes and Practice

Matching bases only works when both sides of an equation happen to be powers of the same number. Most real equations are not that polite: a population reaching 8500, an investment reaching a target balance, a sample decaying to a particular mass. Logarithms remove that restriction completely, because the power law can pull any exponent down where you can reach it.

Exponential equations: take the logarithm of both sides

b^{f (x)} = M ⟹ lo g (b^{f (x)}) = lo g M ⟹ f (x) lo g b = lo g M ⟹ f (x) = \frac{lo g M}{lo g b}

If two quantities are equal then their logarithms are equal, so you may take the log of both whole sides. The power law then drags the variable down out of the exponent, leaving an ordinary linear equation.

When matching bases is still faster

If both sides can be written as powers of the same base, matching bases is quicker and gives an exact answer instead of a decimal. For

8^{x} = 32

, write

2^{3 x} = 2^{5}

and read off

x = \frac{5}{3}

in one line. Reach for logarithms only when no common base exists, as in

8^{x} = 30

.

VideoJensenMath, an Ontario teacher, solves exponential equations both ways: by matching bases when that is possible, and by taking the logarithm of both sides when it is not.Watch on YouTube

Logarithmic equations

A logarithmic equation has the variable inside a logarithm. The strategy is always the same: condense both sides down to a single logarithm, then convert to exponential form and solve whatever algebra is left — usually a linear or quadratic equation.

State the domain first: every argument must be positive, so write down each inequality before you do any algebra.
Use the logarithm laws to condense each side into a single logarithm.
If the equation has the form $lo g_{b} A = lo g_{b} B$ , then $A = B$ . If it has the form $lo g_{b} A = c$ , convert to $A = b^{c}$ .
Solve the resulting equation for all candidate values of the variable.
Test every candidate against the domain from step 1, and reject any that fail. State clearly which roots are extraneous.
Substitute the surviving roots back into the original equation to confirm.

Always check for extraneous roots

Condensing two logarithms into one quietly widens the domain. The original

lo g_{2} x + lo g_{2} (x - 2)

requires

x > 2

, but the condensed

lo g_{2} (x (x - 2))

is happy with any

x

that makes the product positive — including negative values of

x

. So the algebra that follows can hand you answers the original equation never allowed. Those answers are called extraneous roots, and rejecting them explicitly is part of the solution, not an optional check. One subtlety: a negative candidate is not automatically extraneous. What has to come out positive is each argument, not

x

itself — in

lo g (x + 9) = 1

a negative root would be perfectly acceptable as long as it kept

x + 9

above zero. Always test the arguments, never just the sign of

x

.

Logarithmic scales in the real world

Some quantities vary over such huge ranges that a linear scale is useless. The intensity of a barely audible sound and that of a jet engine differ by a factor of more than a trillion. A logarithmic scale compresses that range into manageable numbers by reporting the exponent instead of the quantity — which is why every one of these scales is defined with a logarithm.

Fig 1pH, decibels and the Richter magnitude scale all compress enormous ranges by reporting an exponent instead of a raw value.

Scale	Definition	What one unit means
pH (acidity)	$pH = - lo g [H^{+}]$ , concentration in mol/L	One pH unit is a 10-fold change in hydrogen ion concentration; lower pH means more acidic.
Sound level	$β = 10 lo g (\frac{I}{I _{0}})$ , $I_{0} = 1 0^{- 12}$ W/m²	10 dB is a 10-fold change in intensity; 20 dB is 100-fold.
Richter magnitude	$M = lo g (\frac{I}{I _{0}})$	One magnitude unit is a 10-fold change in wave amplitude.

Working with the pH scale

Orange juice has a hydrogen ion concentration of about

3.2 \times 1 0^{- 5}

mol/L, so its pH is

- lo g (3.2 \times 1 0^{- 5})

. Split it with the product law:

- (lo g 3.2 + lo g 1 0^{- 5}) = - (0.505 - 5) \approx 4.5 .

Going the other way is even easier — a solution of pH 8 has

[H^{+}] = 1 0^{- 8}

mol/L exactly, because the definition just undoes itself.

A paper-covered drum on a seismograph with rows of wavy ink traces, some rows swinging much wider than others. — PhotoA seismograph at Weston Observatory in Massachusetts records ground shaking on a rotating paper drum. Magnitude scales are built from the logarithm of wave amplitudes like these, which is why one step up in magnitude means a tenfold change in amplitude.Z22, CC BY-SA 3.0, via Wikimedia Commons

VideoA full JensenMath lesson for this course applying the logarithm laws to pH, sound intensity in decibels and earthquake magnitudes.Watch on YouTube

Growth, decay and half-life

A (t) = A_{0} (1 + r)^{t}, A (t) = A_{0} (\frac{1}{2})^{t / h}, A (t) = A_{0} (2)^{t / D}

Growth or decay at rate r, half-life h, and doubling time D. These models become solvable for time once you have logarithms: before, you could only ask how much is left after 20 hours; now you can ask how long until only 10 mg is left, and answer it exactly.

Fig 2Growth and decay curves: logarithms let you read the horizontal axis, answering how long rather than how much.

Solving Exponential and Logarithmic Equations

Key things to remember

Exponential equations: take the logarithm of both sides

Logarithmic equations

Logarithmic scales in the real world

Growth, decay and half-life

Worked examples

Where marks get lost

Flashcards 10

Practice