14.6PureStretch

Solving Equations Using Logarithms

When the unknown is stuck in a power, logarithms set it free. Taking logs of both sides and using the power law brings the exponent down where you can solve for it.

30 min Video by Zeeshan Zamurred Exponentials and Logarithms

Edexcel AS Level Maths: 14.6 Solving Equations Using LogarithmsWatch the full walkthrough before the notes below.

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What you'll be able to do

Take logs of both sides of an equation
Use the power law to bring down the exponent
Solve equations of the form aˣ = b
Give answers to a suitable accuracy

Take logs of both sides

For an equation like $a^{x} = b$ , taking logs of both sides and applying the power law turns the exponent into a multiplier you can divide out.

a^{x} = b \Rightarrow x lo g a = lo g b \Rightarrow x = \frac{lo g b}{lo g a}

Logs free the variable from the exponent.

1Take logs:

x lo g 3 = lo g 20

x = \frac{lo g 20}{lo g 3}

Answer

x \approx 2.73

Tip — You can use any base of log (usually log₁₀ or ln) — just be consistent on both sides.

Hidden quadratics in eˣ

Some exponential equations are quadratics in disguise — e.g. in $e^{2 x}$ and $e^{x}$ . Substitute $y = e^{x}$ , solve the quadratic, then take logs to finish.

1Let

y = e^{x}

y^{2} - 5 y + 6 = 0

2Factorise:

(y - 2) (y - 3) = 0

, so

e^{x} = 2

e^{x} = 3

3Take logs:

x = ln 2

x = ln 3

Answer

x = ln 2, ln 3

Formula recap

a^{x} = b \Rightarrow x = \frac{lo g b}{lo g a}

Solve aˣ = b.

x lo g a = lo g b

Power law brings x down.

y = e^{x} substitution

For hidden quadratics.

Common mistakes to avoid

Writing log b / log a as log(b/a).

They are different! log b ÷ log a is a division of two logs, not log(b/a).

Forgetting the variable can be brought down only via the power law.

Take logs of both sides first, then apply log(aˣ) = x log a.

Key takeaways

Take logs of both sides to free a variable from the exponent.
aˣ = b ⟹ x = log b / log a.
Hidden quadratics in eˣ: substitute y = eˣ, solve, then take logs.

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