How to Solve Equations with Logarithms?

Solving equations with logarithms can seem tricky at first, but once you understand the basic principles, it becomes much easier. Let’s break it down step by step.

What is a Logarithm?

A logarithm is the inverse operation to exponentiation. This means that the logarithm of a number is the exponent to which the base must be raised to produce that number. For example, in the equation $text{log}_2 8 = 3$, 2 is the base, 8 is the result, and 3 is the exponent.

Basic Properties of Logarithms

Before diving into solving equations, it’s essential to understand some basic properties of logarithms:

  1. Product Property: $text{log}_b (xy) = text{log}_b x + text{log}_b y$
  2. Quotient Property: $text{log}_b frac{x}{y} = text{log}_b x – text{log}_b y$
  3. Power Property: $text{log}_b (x^y) = y text{log}_b x$
  4. Change of Base Formula: $text{log}_b x = frac{text{log}_k x}{text{log}_k b}$, where $k$ is any positive number.

Solving Simple Logarithmic Equations

Example 1: $text{log}_2 (x) = 3$

To solve for $x$, rewrite the logarithmic equation in its exponential form:
$2^3 = x$
So, $x = 8$

Example 2: $text{log} (x) = 2$

Assuming the base is 10 (common logarithm), rewrite it as:
$10^2 = x$
Thus, $x = 100$

Solving More Complex Logarithmic Equations

Example 3: $text{log}_3 (x) + text{log}_3 (x-2) = 1$

  1. Use the product property: $text{log}_3 [x(x-2)] = 1$
  2. Rewrite in exponential form: $3^1 = x(x-2)$
  3. Solve the quadratic equation: $3 = x^2 – 2x$
    $x^2 – 2x – 3 = 0$
  4. Factorize: $(x-3)(x+1) = 0$
  5. Solutions: $x = 3$ or $x = -1$
  6. Check for extraneous solutions: $x = -1$ is not valid because you can’t take the logarithm of a negative number. So, $x = 3$

Example 4: $2text{log}_5 (x) – text{log}_5 (4) = 1$

  1. Use the power property: $text{log}_5 (x^2) – text{log}_5 (4) = 1$
  2. Use the quotient property: $text{log}_5 frac{x^2}{4} = 1$
  3. Rewrite in exponential form: $5^1 = frac{x^2}{4}$
  4. Solve for $x$: $5 = frac{x^2}{4}$
    $x^2 = 20$
    $x = text{sqrt}(20)$
    $x = 2text{sqrt}(5)$

Conclusion

Solving logarithmic equations involves understanding their properties and converting them into exponential form when necessary. Always check for extraneous solutions, especially when dealing with logarithms, as they can only take positive values.

Happy solving!

Citations

  1. 1. Khan Academy – Logarithms
  2. 2. Purplemath – Solving Logarithmic Equations
  3. 3. Math is Fun – Logarithms

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ