Solving Exponential and Logarithmic Equations

This tutorial explains how to solve exponential and logarithmic equations using fundamental properties. Each example includes a detailed solution and, when appropriate, a verification step. Before starting, it is strongly recommended to review the rules of logarithms and exponentials.

Properties and Rules of Logarithms and Exponentials

The following rules will be used throughout this tutorial:

  1. \(\log_b(UV) = \log_b U + \log_b V\)
  2. \(\log_b\!\left(\frac{U}{V}\right) = \log_b U - \log_b V\)
  3. \(\log_b(U^r) = r\log_b U\)
  4. \(\log_b(b^x) = x\)
  5. \(b^{\log_b x} = x \quad (x>0)\)
  6. \(\log_b U = \log_b V \iff U = V\)
  7. \(a^x a^y = a^{x+y}\)
  8. \(\frac{a^x}{a^y} = a^{x-y}\)

Since logarithmic and exponential functions with the same base are inverses, we have the key relationship

\[ \log_b x = y \iff x = b^y \]

This inverse property allows us to convert logarithmic equations into exponential equations and vice versa.

Examples with Solutions

Example 1

Solve the equation

\[ \ln(x) = 5 \]

Solution

Using the inverse relationship between \(\ln\) and \(e^x\):

\[ x = e^5 \]

Check:

\[ \ln(e^5) = 5 \]

The solution is valid.

Example 2

Find all real solutions to

\[ e^x = 6 \]

Solution

\[ x = \ln(6) \]

Check:

\[ e^{\ln(6)} = 6 \]

Example 3

Solve

\[ \ln(x) + \ln(2) = 3 \]

Solution

\[ \ln(2x) = 3 \] \[ 2x = e^3 \] \[ x = \frac{e^3}{2} \]

Example 4

Find all real solutions to

\[ e^{3x} = -1 \]

Solution

The exponential function has range \((0,+\infty)\). Since the right-hand side is negative, there are no real solutions.

Example 5

Solve and check

\[ \ln(x) + 2 = -3\ln(x) + 10 \]

Solution

\[ 4\ln(x) = 8 \] \[ \ln(x) = 2 \] \[ x = e^2 \]

Check:

\[ \ln(e^2) + 2 = 4 \] \[ -3\ln(e^2) + 10 = 4 \]

Example 6

Solve

\[ 2e^x + e^{-x} = 3 \]

Solution

Multiply both sides by \(e^x\):

\[ 2e^{2x} + 1 = 3e^x \]

Let \(u = e^x\):

\[ 2u^2 - 3u + 1 = 0 \] \[ u = 1 \quad \text{or} \quad u = \frac{1}{2} \]

Returning to \(x\):

\[ x = 0 \quad \text{or} \quad x = -\ln(2) \]

Both solutions satisfy the original equation.

Example 7

Solve

\[ \ln(x+1) + \ln(x) = \ln(2) \]

Solution

\[ \ln(x(x+1)) = \ln(2) \] \[ x^2 + x - 2 = 0 \] \[ x = 1 \quad \text{or} \quad x = -2 \]

Only \(x=1\) is valid since logarithms require positive arguments.

Example 8

Solve

\[ \ln(x+1) - \ln(x) = 2 \]

Solution

\[ \ln\!\left(\frac{x+1}{x}\right) = 2 \] \[ \frac{x+1}{x} = e^2 \] \[ x = \frac{1}{e^2 - 1} \]

Example 9

Solve

\[ e^{2x}e^{3x} - 3 = 2 \]

Solution

\[ e^{5x} = 5 \] \[ x = \frac{\ln(5)}{5} \]

Example 10

Solve

\[ \ln(x^4) + \ln(x^2) - \ln(x^3) - 2 = 7 \]

Solution

\[ \ln(x^3) = 9 \] \[ x^3 = e^9 \] \[ x = e^3 \]

More References