5. Logarithmic equations · Equations and systems

Idea: removing the logarithms

The two key identities

To solve an equation with logarithms, the idea is to get rid of them. There are two ways to do this:

1) Logarithm = logarithm (same base on both sides):

\log_{a}(B) = \log_{a}(C) \;\Longleftrightarrow\; B = C.

2) Logarithm = number (via the inverse definition):

\log_{a}(B) = C \;\Longleftrightarrow\; a^{C} = B.

Tools: the properties of logarithms

To reach one of the two forms above, we often need to combine logarithms using:

\log_{a}(BC) = \log_{a} B + \log_{a} C, \qquad \log_{a}\!\dfrac{B}{C} = \log_{a} B - \log_{a} C, \qquad \log_{a}(B^{n}) = n\,\log_{a} B.

Essential final check

The logarithm is only defined for positive arguments. When we find a candidate solution, we must substitute it back into the original equation and check that no logarithm argument comes out $\le 0$. If one does, it is discarded as an extraneous solution.

Worked example (two ways)

$\dfrac{\log_{3} 9^{x+1}}{x} = 1$

We multiply by $x$ to clear the denominator:

\log_{3} 9^{x+1} = x.

From here, two ways to reach the same result:

Way 1 — inverse definition ($\log_{a} B = C \Leftrightarrow a^{C} = B$):

3^{x} = 9^{x+1} = (3^{2})^{x+1} = 3^{2x+2}.

Equating exponents: $\;x = 2x+2 \Longrightarrow x = -2$.

Way 2 — properties of the logarithm (pulling out exponents):

\log_{3} 9^{x+1} = \log_{3}\bigl(3^{2(x+1)}\bigr) = 2x+2.

So $\;2x+2 = x \Longrightarrow x = -2$.

Check: $x = -2$: $9^{x+1} = 9^{-1} = \tfrac{1}{9}$ and $\log_{3}\tfrac{1}{9} = -2$. Then $\dfrac{-2}{-2} = 1$. ✓

\boxed{\;x = -2.\;}

Exercises

2 Solve the following logarithmic equations

a) $\log_{x} 64 = 3$

By the inverse definition: $\;x^{3} = 64 \Rightarrow x = \sqrt[3]{64} = 4$.

Check: the base must be $>0$ and $\neq 1$. $x = 4$ ✓.

\boxed{\;x = 4.\;}

b) $\log_{2} x + \log_{2}(x+3) = 2$

We add them as the logarithm of a product:

\log_{2}\!\bigl[x(x+3)\bigr] = 2 \;\Longrightarrow\; x(x+3) = 2^{2} = 4.

x^{2}+3x-4 = 0 \;\Longrightarrow\; (x+4)(x-1) = 0.

$x = -4$ or $x = 1$. Check: $x = -4$ gives $\log_{2}(-4)$, undefined → discard. $x = 1$ ✓ ($\log_{2} 1 + \log_{2} 4 = 0+2 = 2$).

\boxed{\;x = 1.\;}

c) $\log(2x-4) - \log(x+2) = 1$

We subtract them as the logarithm of a quotient:

\log\!\dfrac{2x-4}{x+2} = 1 \;\Longrightarrow\; \dfrac{2x-4}{x+2} = 10.

2x-4 = 10(x+2) = 10x+20 \;\Longrightarrow\; -8x = 24 \;\Longrightarrow\; x = -3.

Check: $x=-3$ ⇒ $2x-4 = -10 < 0$, we cannot take $\log$ → discard.

\boxed{\;\text{No solution.}\;}

A teaching case: we find an algebraic value, but it is not valid because the logarithm is not defined. Here we see why we must always check.

d) $\log x + \log 50 = \log 1000$

We add the left-hand side:

\log(50x) = \log 1000 \;\Longrightarrow\; 50x = 1000 \;\Longrightarrow\; x = 20.

Check: $x = 20 > 0$ ✓.

\boxed{\;x = 20.\;}

e) $\log x = 1 + \log(22-x)$

We move $\log(22-x)$ to the left and write $1 = \log 10$:

\log x - \log(22-x) = 1 \;\Longrightarrow\; \log\!\dfrac{x}{22-x} = 1 \;\Longrightarrow\; \dfrac{x}{22-x} = 10.

x = 10(22-x) = 220 - 10x \;\Longrightarrow\; 11x = 220 \;\Longrightarrow\; x = 20.

Check: $x = 20 > 0$ ✓ and $22-x = 2 > 0$ ✓.

\boxed{\;x = 20.\;}

f) $2\log(x+1) - \log 2 = \log(x^{2}-1)$

We apply the properties:

\log(x+1)^{2} - \log 2 = \log(x^{2}-1) \;\Longrightarrow\; \log\!\dfrac{(x+1)^{2}}{2} = \log(x^{2}-1).

\dfrac{(x+1)^{2}}{2} = x^{2}-1 = (x-1)(x+1).

If $x+1 \neq 0$, we can cancel $(x+1)$ on both sides:

\dfrac{x+1}{2} = x-1 \;\Longrightarrow\; x+1 = 2x-2 \;\Longrightarrow\; x = 3.

Check: $x=3$: $x+1=4>0$ ✓, $x^{2}-1=8>0$ ✓. $2\log 4 - \log 2 = \log 16 - \log 2 = \log 8 = \log(9-1)$ ✓.

\boxed{\;x = 3.\;}

g) $\log(3x+1) - \log(2x-3) = 1 - \log 5$

We move $\log 5$ to the left and group:

\log(3x+1) - \log(2x-3) + \log 5 = 1 \;\Longrightarrow\; \log\!\dfrac{5(3x+1)}{2x-3} = 1.

\dfrac{5(3x+1)}{2x-3} = 10 \;\Longrightarrow\; 5(3x+1) = 10(2x-3).

15x+5 = 20x-30 \;\Longrightarrow\; 5x = 35 \;\Longrightarrow\; x = 7.

Check: $3x+1 = 22 > 0$ ✓, $2x-3 = 11 > 0$ ✓.

\boxed{\;x = 7.\;}

h) $2\log(2x+1) + 2\log(3x-4) = 2$

We divide everything by $2$ (or, equivalently, pull out the exponents and use the product property):

\log(2x+1) + \log(3x-4) = 1 \;\Longrightarrow\; \log\!\bigl[(2x+1)(3x-4)\bigr] = 1.

(2x+1)(3x-4) = 10 \;\Longrightarrow\; 6x^{2}-5x-14 = 0.

x = \dfrac{5 \pm \sqrt{25+336}}{12} = \dfrac{5 \pm 19}{12} \;\Rightarrow\; x = 2\ \text{or}\ x = -\tfrac{7}{6}.

Check: $x=2$: $2x+1=5>0$ ✓, $3x-4=2>0$ ✓. $x=-\tfrac{7}{6}$: $2x+1=-\tfrac{4}{3}<0$ → discard.

\boxed{\;x = 2.\;}

i) $\log(40x) - \log(5x-1) = 1$

\log\!\dfrac{40x}{5x-1} = 1 \;\Longrightarrow\; \dfrac{40x}{5x-1} = 10.

40x = 10(5x-1) = 50x-10 \;\Longrightarrow\; -10x = -10 \;\Longrightarrow\; x = 1.

Check: $40x = 40 > 0$ ✓, $5x-1 = 4 > 0$ ✓.

\boxed{\;x = 1.\;}

j) $\log x + \log(x+5) = \log(9+x)$

\log\!\bigl[x(x+5)\bigr] = \log(9+x) \;\Longrightarrow\; x(x+5) = 9+x.

x^{2}+5x = 9+x \;\Longrightarrow\; x^{2}+4x-9 = 0.

x = \dfrac{-4 \pm \sqrt{16+36}}{2} = \dfrac{-4 \pm \sqrt{52}}{2} = -2 \pm \sqrt{13}.

Check: $\sqrt{13} \approx 3.606$. $x = -2+\sqrt{13} \approx 1.606 > 0$ ✓. $x = -2-\sqrt{13} \approx -5.606 < 0$ → discard ($\log x$ undefined).

\boxed{\;x = -2 + \sqrt{13}.\;}

k) $2\log_{2} x = -\log_{2} 3 + \log_{2}(x+10)$

We move $\log_{2} 3$ to the left and group:

\log_{2} x^{2} + \log_{2} 3 = \log_{2}(x+10) \;\Longrightarrow\; \log_{2}(3x^{2}) = \log_{2}(x+10).

3x^{2} = x+10 \;\Longrightarrow\; 3x^{2}-x-10 = 0 \;\Longrightarrow\; x = \dfrac{1 \pm \sqrt{1+120}}{6} = \dfrac{1 \pm 11}{6}.

$x = 2$ or $x = -\tfrac{5}{3}$. Check: $x=2>0$ ✓. $x=-\tfrac{5}{3}<0$ → discard.

\boxed{\;x = 2.\;}