Grade 12 · MHF4U · Advanced Functions

MHF4U: Solving Logarithmic Equations

Almost every log equation on an MHF4U test is one of two types. Condense the logs with the log laws, spot which type you have, and the rest is ordinary algebra.

How It Works

Type 1: a log equals a number. $\log_a(\text{stuff}) = c$ . Rewrite it in exponential form: $\text{stuff} = a^c$ . The base, the result and the exponent just change places.

Type 2: a log equals a log. $\log_a M = \log_a N$ , same base on both sides. Then $M = N$ , and the logs drop out.

Before you can use either, each side needs to be a single log, so condense first with the product, quotient and power laws. A constant mixed in with logs can be turned into a log too: $3 = \log_2 8$ .

Then the step people skip: check every answer in the original equation. A log is only defined for a positive argument, so any candidate that makes an argument zero or negative is extraneous and gets rejected. Condensing hides this, because $\log x + \log(x - 6)$ needs $x > 6$ while $\log(x(x - 6))$ also accepts negative $x$ . The quadratic you end up solving will happily hand you a root the original equation cannot use.

One more trap: $\log_a M + \log_a N = c$ is not $M + N = a^c$ . Combine into $\log_a(MN)$ first, then convert.

If the unknown is in an exponent instead, like $5^x = 24$ , it is an exponential equation. Take the log of both sides and the power law brings $x$ down: $x = \frac{\log 24}{\log 5}$ . The lesson covers both kinds side by side.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Solve $\log_2(x + 3) = 4$ .
Show solution
1. Type 1. Exponential form: $x + 3 = 2^4 = 16$ .
2. So $x = 13$ . Check: $\log_2 16 = 4$ , and the argument is positive.
Answer: $x = 13$
Easy
Solve $\log(3x - 2) = \log(x + 6)$ .
Show solution
1. Type 2, same base on both sides, so the arguments are equal: $3x - 2 = x + 6$ .
2. $2x = 8$ , so $x = 4$ .
3. Check: both arguments are $10$ , positive.
Answer: $x = 4$
Medium
Solve $\log_3 x + \log_3(x - 6) = 3$ .
Show solution
1. Condense with the product law: $\log_3(x(x - 6)) = 3$ .
2. Exponential form: $x(x - 6) = 3^3 = 27$ , so $x^2 - 6x - 27 = 0$ .
3. Factor: $(x - 9)(x + 3) = 0$ , so $x = 9$ or $x = -3$ .
4. Check: $x = -3$ makes $\log_3 x$ undefined, so reject it. $x = 9$ gives $\log_3 9 + \log_3 3 = 2 + 1 = 3$ .
Answer: $x = 9$
Medium
Solve $\log_2(5x + 4) - \log_2(x - 1) = 3$ .
Show solution
1. Condense with the quotient law: $\log_2\left(\frac{5x + 4}{x - 1}\right) = 3$ .
2. Exponential form: $\frac{5x + 4}{x - 1} = 8$ .
3. Multiply out: $5x + 4 = 8x - 8$ , so $3x = 12$ and $x = 4$ .
4. Check: the arguments are $24$ and $3$ , both positive, and $\log_2 8 = 3$ .
Answer: $x = 4$
Test level
Solve $\log_2(x + 1) + \log_2(x - 1) = 3 + \log_2 x$ .
Show solution
1. Turn the constant into a log: $3 = \log_2 8$ , so the right side is $\log_2 8 + \log_2 x = \log_2(8x)$ .
2. Condense the left: $\log_2((x + 1)(x - 1)) = \log_2(x^2 - 1)$ .
3. Type 2: $x^2 - 1 = 8x$ , so $x^2 - 8x - 1 = 0$ .
4. Quadratic formula: $x = \frac{8 \pm \sqrt{64 + 4}}{2} = 4 \pm \sqrt{17}$ .
5. Check: $4 - \sqrt{17} \approx -0.12$ makes $\log_2 x$ undefined, so reject it. $4 + \sqrt{17} \approx 8.12$ keeps all three arguments positive.
Answer: $x = 4 + \sqrt{17}$

Learn It on Martian Lab

Martian Lab teaches this in the Logarithmic & Exponential Equations lesson of its MHF4U Exponential & Logarithmic Functions unit. Watch it, then take the Exponential & Logarithmic Functions practice tests to see if it stuck.

Watch the Logarithmic & Exponential Equations lesson free

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Free Exponential & Logarithmic Functions practice test, with solutions

More Exponential & Logarithmic Functions topics: Laws of Logarithms

More MHF4U topics: Long and Synthetic Division · Vertical and Horizontal Asymptotes · Proving Trig Identities · Solving Trig Equations

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