Simplifying logarithmic expressions, solving logarithmic equations, expressing one variable in terms of another, and writing formulas in logarithmic form and back again.

The laws from the previous lesson turn a sum of logarithms into the logarithm of a product, and a difference into the logarithm of a quotient. Thanks to that, an equation with logarithms to the same base on both sides reduces to an ordinary equation between the numbers inside them. The same swap works the other way round: a formula with multiplication, division and powers can be written as a sum and difference of logarithms, which is often convenient in technical calculations.

Logarithmic equations

Simplify.

\begin{aligned}2\log_{a} x - 3\log_{a}(2x) + \log_{a} x^{2} &= \log_{a} x^{2} - \log_{a}(2x)^{3} + \log_{a} x^{2} = \\&= \log_{a} x^{2} - \log_{a} 8x^{3} + \log_{a} x^{2} = \\&= \log_{a} \frac{x^{2}}{8x^{3}} \cdot x^{2} = \log_{a} \frac{x^{4}}{8x^{3}} = \\&= \log_{a} \frac{x}{8} = \log_{a} \frac{1}{8}x\end{aligned}

Solve the following equation for x.

4\log_{a} \sqrt{x} - \log_{a}(3x) = \log_{a} x^{-2}

\log_{a}\left(x^{\tfrac{1}{2}}\right)^{4} - \log_{a}(3x) = \log_{a} x^{-2}

\log_{a} x^{2} - \log_{a}(3x) = \log_{a} x^{-2}

\displaystyle \frac{x^{2}}{3x} = \frac{1}{x^{2}}

\displaystyle \frac{x}{3} = \frac{1}{x^{2}}

x^{3} = 3

x = \sqrt[3]{3}

Find y in terms of x.

2\log_{a} y - 3\log_{a}(x^{2}) = \log_{a} \sqrt{y} + \log_{a} x

\log_{a} y^{2} - \log_{a}(x^{2})^{3} = \log_{a} \sqrt{y} + \log_{a} x

\log_{a} y^{2} - \log_{a} \sqrt{y} = \log_{a} x + \log_{a} x^{6}

\displaystyle \frac{y^{2}}{\sqrt{y}} = x \cdot x^{6}

\displaystyle \frac{y^{2}}{y^{\tfrac{1}{2}}} = x^{7}

y^{\tfrac{3}{2}} = x^{7}

y = \left(x^{7}\right)^{\tfrac{2}{3}}

y = x^{\tfrac{14}{3}}

y = \sqrt[3]{x^{14}}

Review exercises

Simplify the following expressions.

\text{a)}\quad a^{6} \cdot a^{5} = a^{11}

\text{b)}\quad x^{7} : x^{3} = x^{4}

\text{c)}\quad (w^{2})^{m} : (w^{m})^{3} = w^{2m} : w^{3m} = w^{-m}

\text{d)}\quad s^{3} : t^{-4} \cdot (s^{-3}t^{-2})^{3} = s^{3} : t^{-4} \cdot s^{-9} \cdot t^{-6} = s^{3} \cdot t^{4} \cdot s^{-9} \cdot t^{-6} = s^{-6}t^{-2}

\text{e)}\quad \frac{8x^{-3} \cdot 3x^{2}}{6x^{-4}} = \frac{24x^{-1}}{6x^{-4}} = 4x^{3}

\begin{aligned}\text{f)}\quad (4a^{3}b^{-1}c)^{2} \cdot \left(a^{-2}b^{4}c^{-2}\right)^{\tfrac{1}{2}} : \left[64\left(a^{6}b^{4}c^{2}\right)^{-\tfrac{1}{2}}\right] &= \\&\hspace{-10em} = 16a^{6}b^{-2}c^{2} \cdot a^{-1}b^{2}c^{-1} : \left(64a^{-3}b^{-2}c^{-1}\right) = \\&\hspace{-10em} = \frac{16a^{5}b^{0}c}{64a^{-3}b^{-2}c^{-1}} = \frac{1}{4}a^{8}b^{2}c^{2}\end{aligned}

\begin{aligned}\text{g)}\quad \sqrt[3]{8a^{3}b^{6}} : \sqrt{\frac{1}{25}a^{4}b^{7}} \cdot \left(16\sqrt{a^{4}b^{6}}\right)^{-\tfrac{1}{2}} &= \\&\hspace{-10em} = \left(8a^{3}b^{6}\right)^{\tfrac{1}{3}} : \left(\frac{1}{25}a^{4}b^{7}\right)^{\tfrac{1}{2}} \cdot \left(16a^{2}b^{3}\right)^{-\tfrac{1}{2}} = \\&\hspace{-10em} = 2ab^{2} : \left(\frac{1}{5}a^{2}b^{\tfrac{7}{2}}\right) \cdot \frac{1}{4}a^{-1}b^{-\tfrac{3}{2}} = \\&\hspace{-10em} = 2ab^{2} \cdot 5a^{-2}b^{-\tfrac{7}{2}} \cdot \frac{1}{4}a^{-1}b^{-\tfrac{3}{2}} = \frac{5}{2}a^{-2}b^{-3}\end{aligned}

Rewrite the following expressions without using logarithms.

\begin{aligned}\text{a)}\quad \log K &= \log P - \log T + 1.3\log V \\\log K &= \log \frac{P}{T} + \log V^{1.3} \\\log K &= \log \frac{PV^{1.3}}{T} \quad \longrightarrow \quad K = \frac{PV^{1.3}}{T}\end{aligned}

\begin{aligned}\text{b)}\quad \ln A &= \ln P + m \\m &= \ln A - \ln P \\m &= \ln \frac{A}{P} \quad \longrightarrow \quad \frac{A}{P} = e^{m} \;/ \cdot P \quad \longrightarrow \quad A = Pe^{m}\end{aligned}

Write R in logarithmic form.

R = r\sqrt{\frac{f + P}{f - P}}

\begin{aligned}R &= r\frac{\sqrt{f + P}}{\sqrt{f - P}} \quad \longrightarrow \quad \log R = \log\left(r\frac{\sqrt{f + P}}{\sqrt{f - P}}\right) \\\log R &= \log r + \log\sqrt{f + P} - \log\sqrt{f - P}\end{aligned}

Evaluate using a calculator or, where necessary, by applying the change-of-base formula (correct to three decimal places).

\text{a)}\quad \log 5.324 = 0.726

\text{b)}\quad \ln 0.0023 = -6.075

\text{c)}\quad \log_{4} 1.2 = 0.132

\log_{4} 10 \cdot \log_{10} 1.2 = \frac{\log_{10} 1.2}{\log_{10} 4} = \frac{0.079}{0.602} = 0.132

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Frequently asked questions

How do you solve an equation with logarithms on both sides?

First use the laws to reduce each side to a single logarithm, then compare the numbers inside them. From log_a x² − log_a 3x = log_a x⁻² you get x²/3x = 1/x², that is x³ = 3.

How do you rewrite a formula without logarithms?

Combine the right-hand side into a single logarithm and then drop the logarithm on both sides. From log K = log P − log T + 1.3 log V you get the logarithm of a quotient whose numerator is P times V to the power 1.3 and whose denominator is T. Hence K equals that same quotient.

How do you get rid of a natural logarithm in an equation?

By raising e to the power of both sides. From m = ln(A/P) it follows that A divided by P equals e to the power m, and multiplying by P gives A equal to P times e to the power m.

Why write a formula in logarithmic form?

Because multiplication and division turn into addition and subtraction, and roots and powers into multiplication by a number. The formula becomes easier to evaluate and to rearrange for any of its variables.

A lesson from the PhiBoard knowledge base: phiboard.com/en/knowledge-base/from-zero-to-engineer/logarithmic-equations. On the lesson page you will also find games and practice exercises - free and without an account.

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Logarithms: definition, laws and change of base

Multiplication and division of algebraic expressions