Lesson 6.6 : Exponential and Logarithmic Equations

New Concept

This lesson covers algebraic techniques for solving exponential and logarithmic equations. You'll learn to find unknown variables by using common bases, applying logarithms, and using one-to-one properties—essential skills for modeling real-world growth and decay.

What’s next

Now, let's put this into practice. You'll start with interactive examples on solving exponential equations by finding like bases, a key first step.

Property

For any algebraic expressions $S$ and $T$, and any positive real number $b \neq 1$,

To solve, use the rules of exponents to simplify, if necessary, so that the resulting equation has the form $b^S = b^T$. Then, use the one-to-one property to set the exponents equal, $S=T$, and solve for the unknown.

Examples

• Solve $7^{x-3} = 7^{2x-9}$. By the one-to-one property, the exponents must be equal: $x-3 = 2x-9$. Solving for $x$ gives $x=6$.

• Solve $9^{x+1} = 27^x$. First, rewrite both sides with a common base of 3: $(3^2)^{x+1} = (3^3)^x$. This simplifies to $3^{2x+2} = 3^{3x}$. Now, set the exponents equal: $2x+2 = 3x$, so $x=2$.

Property

Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Since $\operatorname{log}(a) = \operatorname{log}(b)$ is equivalent to $a = b$, we may apply logarithms with the same base on both sides of an exponential equation.

To solve, apply the logarithm of both sides of the equation. Then, use the rules of logarithms to solve for the unknown.

Examples

• Solve $3^{x+1} = 5^x$. Take the natural log of both sides: $\ln(3^{x+1}) = \ln(5^x)$. The power rule gives $(x+1)\ln(3) = x\ln(5)$. Solving for $x$ yields $x = \frac{\ln(3)}{\ln(5)-\ln(3)}$.

Property

When we have an equation with a base $e$ on either side, we can use the natural logarithm to solve it.

To solve an equation of the form $y = Ae^{kt}$:

1. Divide both sides of the equation by $A$ to isolate the exponential expression.
2. Apply the natural logarithm of both sides of the equation.
3. Divide both sides of the equation by $k$ to solve for $t$.

Examples

• Solve $250 = 50e^{2t}$. First, divide by 50 to get $5 = e^{2t}$. Take the natural log of both sides: $\ln(5) = \ln(e^{2t})$, which simplifies to $\ln(5) = 2t$. Thus, $t = \frac{\ln(5)}{2}$.

Property

Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises when solving logarithmic equations.
In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative or zero, there is no output.

Examples

• Solve $\operatorname{log}(x) + \operatorname{log}(x-3) = \operatorname{log}(4)$. Combine logs to get $\operatorname{log}(x(x-3)) = \operatorname{log}(4)$, so $x^2-3x-4=0$. Factoring gives $(x-4)(x+1)=0$. Solutions are $x=4$ and $x=-1$. Only $x=4$ is valid, as $x=-1$ makes the original arguments negative.

• Solve $e^{2x} - 2e^x - 15 = 0$. Let $u=e^x$, so $u^2 - 2u - 15 = 0$, which factors to $(u-5)(u+3) = 0$. This gives $e^x=5$ or $e^x=-3$. Since $e^x$ must be positive, $e^x=-3$ is impossible. The only solution is $x=\ln(5)$.

Property

For any algebraic expression $S$ and real numbers $b$ and $c$, where $b > 0, b \neq 1$,

This property allows you to convert between logarithmic and exponential forms.

Examples

• Solve $4\ln x + 1 = 9$. First, isolate the logarithm: $4\ln x = 8$, so $\ln x = 2$. Using the definition (which implies base $e$), this becomes $x = e^2$.

• Solve $\operatorname{log}_3(x-4) = 2$. Rewriting in exponential form gives $3^2 = x-4$. This simplifies to $9 = x-4$, so the solution is $x=13$.

Property

For any algebraic expressions $S$ and $T$ and any positive real number $b$, where $b \neq 1$,

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

Examples

• Solve $\ln(x^2 - 2) = \ln(2x+1)$. Using the one-to-one property, set the arguments equal: $x^2 - 2 = 2x+1$. This gives $x^2-2x-3=0$, which factors to $(x-3)(x+1)=0$. The solutions are $x=3$ and $x=-1$.

• Solve $\operatorname{log}_5(4x-3) = \operatorname{log}_5(x+6)$. Since the bases are the same, set the arguments equal: $4x-3 = x+6$. This simplifies to $3x = 9$, so $x=3$. Checking the solution shows the arguments are positive.