Solving logarithmic equations

A log­a­rith­mic equa­tion has the unknown inside a log­a­rithm, or in the base of one. Solv­ing it uses two tools: the laws, to col­lect every­thing into a sin­gle log­a­rithm, and the def­i­n­i­tion, to remove the log­a­rithm alto­gether. There is one extra step that ordi­nary equa­tions do not need: every answer must be checked, because a log­a­rithm only exists for a pos­i­tive num­ber and an allowed base. Answers that fail the check are thrown away.

The tools#

Remove the log­a­rithm using the def­i­n­i­tion:

$$\log_a N = x \implies N = a^x$$

Two spe­cial cases come up again and again: $\log_a N = 0$ means $N = 1$, and $\log_a N = 1$ means $N = a$.

Remove log­a­rithms from both sides (Law 12): $\log_a m = \log_a n \implies m = n$.

Col­lect log­a­rithms first: $\log_a m + \log_a n = \log_a mn$ and $\displaystyle \log_a m - \log_a n = \log_a \frac{m}{n}$.

Check every answer in the orig­i­nal equa­tion. Each num­ber inside a log­a­rithm must be pos­i­tive, and each base must be pos­i­tive and not 1.

Exam­ple 1#

If $\log_5(2x + 3) = 0$, find $x$.

A log­a­rithm is 0 only when the num­ber is 1, so $2x + 3 = 1$. Then $2x = -2$ and $x = -1$. Check: $2(-1) + 3 = 1$, which is pos­i­tive. So $x = -1$.

Exam­ple 2#

If $\log_5(2x - 3) = 1$, find $x$.

A log­a­rithm is 1 only when the num­ber equals the base, so $2x - 3 = 5$. Then $2x = 8$ and $x = 4$. Check: $2(4) - 3 = 5 > 0$.

Exam­ple 3#

If $\log_2(x^2 + 5x + 8) = 1$, find $x$.

The num­ber must equal the base: $x^2 + 5x + 8 = 2$, so $x^2 + 5x + 6 = 0$. Split the mid­dle term:

$$\begin{aligned} x^2 + 2x + 3x + 6 &= 0 \\ x(x + 2) + 3(x + 2) &= 0 \\ (x + 2)(x + 3) &= 0 \end{aligned}$$

So $x = -2$ or $x = -3$. Check both: for $x = -2$, $4 - 10 + 8 = 2$; for $x = -3$, $9 - 15 + 8 = 2$. Both make the num­ber 2, which is pos­i­tive, so both answers stand. A neg­a­tive value of $x$ is per­fectly fine; it is the num­ber inside the log­a­rithm that must be pos­i­tive.

Exam­ple 4#

If $\log_2 x + \log_2(x + 1) = 1$, find $x$.

Com­bine with the prod­uct law: $\log_2 x(x + 1) = 1$, that is $\log_2(x^2 + x) = 1$. So

$$\begin{aligned} x^2 + x &= 2 \\ x^2 + x - 2 &= 0 \\ x^2 + 2x - x - 2 &= 0 \\ x(x + 2) - 1(x + 2) &= 0 \\ (x - 1)(x + 2) &= 0 \end{aligned}$$

So $x = 1$ or $x = -2$. Now check. If $x = -2$, the orig­i­nal equa­tion con­tains $\log_2(-2)$, which is not defined. So $x = -2$ is rejected. For $x = 1$: $\log_2 1 + \log_2 2 = 0 + 1 = 1$. The answer is $x = 1$.

Exam­ple 5#

If $\log_{10}(x - 9) + \log_{10} x = 1$, find $x$.

$$\begin{aligned} \log_{10} x(x - 9) &= 1 \\ x^2 - 9x &= 10 \\ x^2 - 9x - 10 &= 0 \\ x^2 - 10x + x - 10 &= 0 \\ (x - 10)(x + 1) &= 0 \end{aligned}$$

So $x = 10$ or $x = -1$. For $x = -1$, $\log_{10}(-1)$ is not defined, so it is rejected. The answer is $x = 10$. Check: $\log_{10} 1 + \log_{10} 10 = 0 + 1 = 1$.

Exam­ple 6#

If $\log_2 x - \log_2(x + 1) = 5$, find $x$.

Use the quo­tient law: $\displaystyle \log_2 \frac{x}{x + 1} = 5$, so $\displaystyle \frac{x}{x + 1} = 2^5 = 32$. Then

$$\displaystyle x = 32x + 32 \implies -31x = 32 \implies x = -\frac{32}{31}$$

But for this value $\displaystyle \log_2 x = \log_2\left(-\frac{32}{31}\right)$ is not defined. So the equa­tion has no real solu­tion. This makes sense: for pos­i­tive $x$, the frac­tion $\displaystyle \frac{x}{x+1}$ is always less than 1, so its log­a­rithm to the base 2 is neg­a­tive and can never be 5.

Exam­ple 7#

If $\log_2(x^2 + x + 1) + \log_2(x - 1) = 3$, find $x$.

Com­bine: $\log_2 (x^2 + x + 1)(x - 1) = 3$. The iden­tity $(a - b)(a^2 + ab + b^2) = a^3 - b^3$ gives $(x - 1)(x^2 + x + 1) = x^3 - 1$. So

$$x^3 - 1 = 2^3 = 8 \implies x^3 = 9 \implies x = \sqrt[3]{9}$$

Check: $\sqrt[3]{9}$ is about 2.08, so $x - 1 > 0$ and $x^2 + x + 1 > 0$. The answer is $x = \sqrt[3]{9}$.

Exam­ple 8#

Solve $\log_2(3x - 2) = \log_{1/2} x$.

Make the bases the same. Since $\displaystyle \frac{1}{2} = 2^{-1}$, Law 5 gives $\displaystyle \log_{2^{-1}} x = \frac{1}{-1} \log_2 x = \log_2 x^{-1}$. So

$$\displaystyle \begin{aligned} \log_2(3x - 2) &= \log_2 x^{-1} \\ 3x - 2 &= \frac{1}{x} \\ 3x^2 - 2x - 1 &= 0 \\ 3x^2 - 3x + x - 1 &= 0 \\ 3x(x - 1) + 1(x - 1) &= 0 \\ (3x + 1)(x - 1) &= 0 \end{aligned}$$

So $x = 1$ or $\displaystyle x = -\frac{1}{3}$. For $\displaystyle x = -\frac{1}{3}$, $\log_{1/2} x$ is not defined. The answer is $x = 1$. Check: $\log_2 1 = 0$ and $\log_{1/2} 1 = 0$.

Exam­ple 9#

If $\log_2 \log_3 \log_5 x = 0$, find $x$.

Peel off one log­a­rithm at a time. A log­a­rithm is 0 only when its num­ber is 1, so $\log_3 \log_5 x = 1$. A log­a­rithm is 1 only when its num­ber equals the base, so $\log_5 x = 3$. Finally $x = 5^3 = 125$.

Check from the inside: $\log_5 125 = 3$, $\log_3 3 = 1$, $\log_2 1 = 0$.

When the unknown is in the base#

The same def­i­n­i­tion works. Remem­ber that the base must be pos­i­tive and not 1.

Exam­ple 10#

If $\displaystyle \log_x 4 = -\frac{1}{3}$, find $x$.

In expo­nen­tial form, $4 = x^{-1/3}$. Raise both sides to the power $-3$: $4^{-3} = x$, so $\displaystyle x = \frac{1}{4^3} = \frac{1}{64}$. This is a pos­i­tive base other than 1, so it is accepted.

Exam­ple 11#

If $\log_{1/x} 4 = -2$, find $x$.

In expo­nen­tial form, $\displaystyle 4 = \left(\frac{1}{x}\right)^{-2} = (x^{-1})^{-2} = x^2$. So $x = 2$ or $x = -2$. If $x = -2$, the base $\displaystyle \frac{1}{x} = -\frac{1}{2}$ is neg­a­tive, which is not allowed. So $x = 2$.

Exam­ple 12#

If $\log_5 \sqrt[3]{x} = \log_5 27$, find $x$.

By Law 12, $\sqrt[3]{x} = 27$. Cube both sides: $x = 27^3 = (3^3)^3 = 3^9 = 19683$.

Prac­tice#

1. If $\log_3(4x - 7) = 0$, find $x$.
2. If $\log_{11}(2x + 1) = 1$, find $x$.
3. If $\log_6(x^2 - 5x + 12) = 1$, find $x$.
4. If $\log_5 x = \log_5(3 - x)$, find $x$.
5. If $\displaystyle \log_5 \frac{x - 1}{x - 2} = 1$, find $x$.
6. If $\log_{10} x + \log_{10}(x - 3) = 1$, find $x$.
7. If $\log_3 x - \log_3(x - 2) = 1$, find $x$.
8. If $\displaystyle \log_x 9 = -\frac{2}{3}$, find $x$.
9. If $\log_2 \log_4 x = 1$, find $x$.
10. Solve $\displaystyle \log_3(x + 6) = \log_{1/3} \frac{1}{x}$.

Answers#