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, where appropriate, a verification step.
Before starting, review the rules of logarithms and exponentials.
Properties and Rules of Logarithms and Exponentials
The following rules will be used throughout this tutorial:
- $\log_b(UV) = \log_b U + \log_b V$
- $\log_b\!\left(\frac{U}{V}\right) = \log_b U - \log_b V$
- $\log_b(U^r) = r\log_b U$
- $\log_b(b^x) = x$
- $b^{\log_b x} = x \quad (x>0)$
- $\log_b U = \log_b V \iff U = V$
- $a^x a^y = a^{x+y}$
- $\frac{a^x}{a^y} = a^{x-y}$
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.
Example 1: Solve a Basic Natural Logarithm Equation
Solve the equation:
$$ \ln(x) = 5 $$View Solution
Using the inverse relationship between $\ln$ (base $e$) and $e^x$, convert the equation to its exponential form:
$$ x = e^5 $$Substitute $x = e^5$ back into the original equation:
$$ \ln(e^5) = 5 $$Conclusion: The solution is valid.
Example 2: Solve an Exponential Equation by Taking ln
Find all real solutions to:
$$ e^x = 6 $$View Solution
Take the natural logarithm ($\ln$) of both sides to isolate $x$:
$$ \ln(e^x) = \ln(6) $$ $$ x = \ln(6) $$Substitute $x = \ln(6)$ back into the original equation:
$$ e^{\ln(6)} = 6 $$Conclusion: The solution is valid.
Example 3: Use the Product Rule of Logarithms
Solve the equation:
$$ \ln(x) + \ln(2) = 3 $$View Solution
- Apply the product rule for logarithms ($\ln(A) + \ln(B) = \ln(AB)$): $$ \ln(2x) = 3 $$
- Convert to exponential form: $$ 2x = e^3 $$
- Solve for $x$: $$ x = \frac{e^3}{2} $$
$x = e^3/2 \approx 10.04 > 0$, so both $\ln(x)$ and $\ln(2x)$ are defined.
Conclusion: $x = \dfrac{e^3}{2}$.
Example 4: An Exponential Equation with No Real Solution
Find all real solutions to:
$$ e^{3x} = -1 $$View Solution
The exponential function $e^{y}$ has a range of strictly $(0,+\infty)$. Since the right-hand side is negative, an exponential function can never equal $-1$.
Conclusion: There are no real solutions.
Example 5: Combine Like ln Terms and Verify
Solve and check:
$$ \ln(x) + 2 = -3\ln(x) + 10 $$View Solution
- Collect all $\ln(x)$ terms on the left and constants on the right: $$ \ln(x) + 3\ln(x) = 10 - 2 $$ $$ 4\ln(x) = 8 $$
- Divide by 4: $$ \ln(x) = 2 $$
- Convert to exponential form: $$ x = e^2 $$
Substitute $x = e^2$ back into the original equation:
Left side: $\ln(e^2) + 2 = 2 + 2 = 4$
Right side: $-3\ln(e^2) + 10 = -3(2) + 10 = -6 + 10 = 4$
Conclusion: Both sides are equal. The solution is valid.
Example 6: Quadratic in Form — Substituting u = e^x
Solve the equation:
$$ 2e^x + e^{-x} = 3 $$View Solution
- Multiply the entire equation by $e^x$ to eliminate the negative exponent: $$ 2e^{2x} + 1 = 3e^x $$
- Rearrange into a quadratic-like structure: $$ 2e^{2x} - 3e^x + 1 = 0 $$
- Let $u = e^x$. The equation becomes a standard quadratic: $$ 2u^2 - 3u + 1 = 0 $$
- Factor the quadratic: $$ (2u - 1)(u - 1) = 0 $$ $$ u = \frac{1}{2} \quad \text{or} \quad u = 1 $$
-
Return to $x$ ($e^x = u$) and solve for each:
If $e^x = 1$, then $x = \ln(1) = 0$.
If $e^x = \frac{1}{2}$, then $x = \ln\left(\frac{1}{2}\right) = -\ln(2)$.
Conclusion: Both solutions $x = 0$ and $x = -\ln(2)$ satisfy the original equation.
Example 7: Logarithmic Equation with an Extraneous Solution
Solve the equation:
$$ \ln(x+1) + \ln(x) = \ln(2) $$View Solution
- Apply the product rule for logarithms on the left side: $$ \ln(x(x+1)) = \ln(2) $$
- Since the bases are equal, equate the arguments: $$ x(x+1) = 2 $$ $$ x^2 + x - 2 = 0 $$
- Factor the quadratic: $$ (x+2)(x-1) = 0 $$ $$ x = -2 \quad \text{or} \quad x = 1 $$
Logarithms require strictly positive arguments.
- If $x = -2$, then $\ln(x)$ becomes $\ln(-2)$, which is undefined. This is an extraneous solution.
- If $x = 1$, then arguments $(1+1)=2$ and $(1)=1$ are both positive and valid.
Conclusion: The only valid solution is $x = 1$.
Example 8: Use the Quotient Rule of Logarithms
Solve the equation:
$$ \ln(x+1) - \ln(x) = 2 $$View Solution
- Apply the quotient rule for logarithms: $$ \ln\!\left(\frac{x+1}{x}\right) = 2 $$
- Convert to exponential form: $$ \frac{x+1}{x} = e^2 $$
- Multiply both sides by $x$ to isolate the variable: $$ x + 1 = x e^2 $$
- Group $x$ terms together on one side: $$ 1 = x e^2 - x $$ $$ 1 = x(e^2 - 1) $$
- Solve for $x$: $$ x = \frac{1}{e^2 - 1} $$
Since $e^2 - 1 \approx 6.39 > 0$, we have $x \approx 0.156 > 0$. Both $x$ and $x+1$ are positive, so the solution is valid.
Conclusion: $x = \dfrac{1}{e^2 - 1}$.
Example 9: Use the Product Rule of Exponents
Solve the equation:
$$ e^{2x}e^{3x} - 3 = 2 $$View Solution
- Use exponent rules ($a^m a^n = a^{m+n}$) and add 3 to both sides: $$ e^{5x} = 5 $$
- Convert to logarithmic form: $$ 5x = \ln(5) $$
- Divide by 5: $$ x = \frac{\ln(5)}{5} $$
$x = \ln(5)/5 \approx 0.322$, so $e^{5x} = e^{\ln 5} = 5$, and $5 - 3 = 2$. ✓
Conclusion: $x = \dfrac{\ln(5)}{5}$.
Example 10: Combine Product, Quotient, and Power Rules
Solve the equation:
$$ \ln(x^4) + \ln(x^2) - \ln(x^3) - 2 = 7 $$View Solution
- Move the constant to the right side and apply logarithm properties to combine the $\ln$ terms ($\ln(A) + \ln(B) - \ln(C) = \ln(AB/C)$): $$ \ln\!\left(\frac{x^4 \cdot x^2}{x^3}\right) = 9 $$
- Simplify the exponent algebra inside the argument ($x^6 / x^3 = x^3$): $$ \ln(x^3) = 9 $$
- Bring the exponent down using the power rule ($\ln(x^r) = r\ln(x)$): $$ 3\ln(x) = 9 $$
- Divide by 3: $$ \ln(x) = 3 $$
- Convert to exponential form: $$ x = e^3 $$
$x = e^3 \approx 20.09 > 0$, so all arguments $x^4, x^2, x^3$ are positive.
Conclusion: $x = e^3$.
Frequently Asked Questions
How do you solve an exponential equation?
To solve an exponential equation such as $e^x = 6$, take the logarithm of both sides using the same base as the exponential, then use the inverse property $\log_b(b^x) = x$ to isolate the variable. If the equation is quadratic in form, substitute $u = e^x$ first.
How do you solve a logarithmic equation?
Combine logarithms on each side using the product, quotient, or power rules, then rewrite the equation in exponential form using $\log_b(x) = y \iff x = b^y$. Always check that every logarithmic argument is strictly positive.
Why can an exponential equation have no solution?
Because $e^x$ is always positive, an equation such as $e^{3x} = -1$ has no real solution. The range of the exponential function is $(0, +\infty)$, so it can never equal a negative number.
Why do logarithmic equations produce extraneous solutions?
Combining logarithms can extend the effective domain of the equation. A candidate that solves the transformed equation may make an original logarithmic argument zero or negative, which is undefined. Always verify candidates in the original equation.