How to Solve a Logarithmic Equation?

Solving a logarithmic equation can seem challenging at first, but with a systematic approach, it becomes much easier. Let’s break down the steps with a clear example.

Step-by-Step Guide

  1. Isolate the Logarithm
    First, try to isolate the logarithmic term on one side of the equation. For example, consider the equation:

    $log_2(x) + 3 = 5$

    Subtract 3 from both sides to isolate the logarithm:

    $log_2(x) = 2$

  1. Rewrite the Equation in Exponential Form
    The key to solving logarithmic equations is to rewrite them in their exponential form. Remember, if $log_b(a) = c$, then $b^c = a$. Using this rule, we can rewrite the equation:

    $2^2 = x$

  1. Solve for the Variable
    Now, solve for the variable. In this case, it’s straightforward:

    $x = 4$

  1. Check Your Solution
    Always check your solution by substituting it back into the original equation to ensure it satisfies the equation:

    $log_2(4) + 3 = 5$

    Since $log_2(4) = 2$, the equation holds true:

    $2 + 3 = 5$

Example with Multiple Logarithms

Let’s consider a more complex example:

$log_3(x) + log_3(x-2) = 2$

  1. Combine the Logarithms
    Use the property of logarithms that states $log_b(m) + log_b(n) = log_b(mn)$:

    $log_3(x(x-2)) = 2$

  1. Rewrite in Exponential Form
    Convert the logarithmic equation to its exponential form:

    $3^2 = x(x-2)$

    Simplify and solve the quadratic equation:

    $9 = x^2 – 2x$

    Rearrange to form a standard quadratic equation:

    $x^2 – 2x – 9 = 0$

  1. Solve the Quadratic Equation
    Use the quadratic formula $x = frac{-b pm sqrt{b^2 – 4ac}}{2a}$, where $a = 1$, $b = -2$, and $c = -9$:

    $x = frac{2 pm sqrt{4 + 36}}{2}$

    $x = frac{2 pm 4sqrt{10}}{2}$

    $x = 1 pm 2sqrt{10}$

  1. Check for Extraneous Solutions
    Since logarithms are only defined for positive numbers, check which solutions are valid. Here, only the positive solution works:

    $x = 1 + 2sqrt{10}$

Conclusion

Solving logarithmic equations involves isolating the logarithm, converting it to exponential form, and solving for the variable. Always check your solutions to ensure they are valid within the domain of the original logarithmic function. With practice, these steps will become second nature, and you’ll find solving logarithmic equations much more manageable.

Citations

  1. 1. Khan Academy – Solving Logarithmic Equations
  2. 2. Purplemath – Solarithms: Solving Logarithmic Equations
  3. 3. Mathway – Solving Logarithmic Equations

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