Lesson 10.5: Solve Exponential and Logarithmic Equations

New Concept

Learn to solve equations where variables are in exponents or inside logarithms. By using logarithmic properties and applying logarithms to both sides, you can tackle complex exponential equations and real-world problems involving growth and decay.

What’s next

This card is your foundation. Next, you'll master these skills through interactive examples, practice problems, and real-world application challenges on interest and growth.

Property

For $M > 0$, $N > 0$, $a > 0$, and $a \neq 1$ is any real number:

Examples

• Solve $2\log_3 x = \log_3 49$. Using the Power Property, this becomes $\log_3 x^2 = \log_3 49$. By the One-to-One Property, $x^2 = 49$, so $x=7$. We discard $x=-7$ since the log of a negative number is undefined.

• Solve $\log_2 (3x - 1) = \log_2 (x + 5)$. The One-to-One Property gives $3x - 1 = x + 5$. Solving for $x$ yields $2x = 6$, so $x=3$. Checking the original equation shows this is a valid solution.

Property

To solve logarithmic equations, you can condense sums or differences into a single logarithm using the Product Property, $\log_a M + \log_a N = \log_a (M \cdot N)$, or the Quotient Property. Once condensed, rewrite the equation in exponential form to solve.

Examples

• Solve $\log_2 x + \log_2 (x - 4) = 5$. Condensing gives $\log_2 [x(x - 4)] = 5$. In exponential form, $x(x-4) = 2^5 = 32$. This gives $x^2 - 4x - 32 = 0$, which factors to $(x-8)(x+4)=0$. The valid solution is $x=8$.

• Solve $\log (x + 15) - \log x = 1$. Using the Quotient Property, $\log \frac{x+15}{x} = 1$. In exponential form, $\frac{x+15}{x} = 10^1 = 10$. Solving for $x$ gives $x+15 = 10x$, so $9x=15$, and $x = \frac{5}{3}$.

Property

When it is not possible to write the expressions with the same base, take the common logarithm or natural logarithm of both sides once the exponential is isolated. Then use the Power Property to move the exponent to the front as a factor and solve for the variable.

Examples

• Solve $3^x = 20$. Take the common log of both sides: $\log(3^x) = \log(20)$. Using the Power Property, $x \log 3 = \log 20$. So, $x = \frac{\log 20}{\log 3} \approx 2.727$.

• Solve $4e^{x-1} = 30$. First, isolate the exponential: $e^{x-1} = 7.5$. Take the natural log of both sides: $\ln(e^{x-1}) = \ln(7.5)$. This simplifies to $x-1 = \ln(7.5)$, so $x = \ln(7.5) + 1 \approx 3.015$.

Property

For a principal, $P$, invested at an interest rate, $r$, for $t$ years, the new balance, $A$ is:

Examples

• How long will it take for 5,000 dollars to grow to 8,000 dollars in an account with 4% interest compounded continuously? We solve $8000 = 5000e^{0.04t}$. This simplifies to $1.6 = e^{0.04t}$, so $t = \frac{\ln(1.6)}{0.04} \approx 11.75$ years.

• An investment of 15,000 dollars needs to grow to 75,000 dollars in 20 years. What continuously compounded interest rate is needed? We solve $75000 = 15000e^{20r}$, which gives $5 = e^{20r}$. Thus, $r = \frac{\ln 5}{20} \approx 0.0805$, or 8.05%.

Property

For an original amount, $A_0$, that grows or decays at a rate, $k$, for a certain time, $t$, the final amount, $A$, is:

Exponential growth has a positive rate of growth or growth constant, $k$, and exponential decay has a negative rate of growth or decay constant, $k$.

Examples

• A yeast culture grows from 50 to 450 cells in 4 hours. To find the growth constant $k$, we solve $450 = 50e^{4k}$. This simplifies to $9 = e^{4k}$, so the constant is $k = \frac{\ln 9}{4} \approx 0.549$.

• The half-life of Sodium-24 is 15 hours. How much of a 20g sample is left after 10 hours? First, find $k$ from $0.5 = e^{15k}$, so $k = \frac{\ln 0.5}{15}$. Then, calculate $A = 20e^{(\frac{\ln 0.5}{15}) \cdot 10} \approx 12.6$ grams.