10.5 Solve Exponential and Logarithmic Equations

Properties and Applications of Logarithms

Solving exponential and logarithmic equations comes down to one core idea: logarithms and exponents undo each other. When a variable is stuck in an exponent, logarithms bring it down. When a variable is trapped inside a logarithm, exponentiation gets it out. This section covers the properties, techniques, and applications you need to solve both types of equations.

Properties of Logarithmic Equations

A logarithm answers the question: "What exponent do I need?" The definition is:

$\log_b(x) = y$ means $b^y = x$

Here, $b$ is the base, $x$ is the argument, and $y$ is the exponent you're solving for. For example, $\log_2(8) = 3$ because $2^3 = 8$.

Three properties let you rewrite and simplify logarithmic expressions:

• Product property: $\log_b(M \cdot N) = \log_b(M) + \log_b(N)$ Multiplication inside the log becomes addition outside. For example, $\log_2(4 \cdot 8) = \log_2(4) + \log_2(8) = 2 + 3 = 5$.

• Quotient property: $\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)$ Division inside the log becomes subtraction outside. For example, $\log_3\left(\frac{27}{9}\right) = \log_3(27) - \log_3(9) = 3 - 2 = 1$.

• Power property: $\log_b(M^n) = n \cdot \log_b(M)$ An exponent inside the log comes out front as a multiplier. For example, $\log_5(25^2) = 2 \cdot \log_5(25) = 2 \cdot 2 = 4$.

The change of base formula lets you evaluate any logarithm on a calculator:

$\log_b(x) = \frac{\log(x)}{\log(b)}$

You can use either common log ($\log$, base 10) or natural log ($\ln$, base $e$) in the formula, as long as you use the same one in both the numerator and denominator. For example, $\log_3(5) = \frac{\log(5)}{\log(3)} \approx \frac{0.699}{0.477} \approx 1.465$.

Solving Exponential Equations with Logarithms

The exponential function $f(x) = b^x$ and the logarithmic function $g(x) = \log_b(x)$ are inverses of each other. This means they cancel each other out, which is exactly what makes logarithms useful for solving exponential equations.

When both sides can be written with the same base, you can solve by matching exponents directly:

1. Rewrite each side as a power of the same base.
2. Set the exponents equal.
3. Solve the resulting equation.

Example: Solve $3^{2x-1} = 27$.

1. Recognize that $27 = 3^3$.

2. So $3^{2x-1} = 3^3$, which means $2x - 1 = 3$.

3. Solve: $2x = 4$, so $x = 2$.

When you can't match bases, take the logarithm of both sides:

1. Take $\log$ or $\ln$ of both sides.
2. Use the power property to bring the variable exponent down.
3. Solve for the variable.

Example: Solve $5^x = 40$.

1. Take the natural log of both sides: $\ln(5^x) = \ln(40)$.
2. Power property: $x \cdot \ln(5) = \ln(40)$.
3. Divide: $x = \frac{\ln(40)}{\ln(5)} \approx \frac{3.689}{1.609} \approx 2.292$.

Solving Logarithmic Equations

To solve equations where the variable is inside a logarithm, convert back to exponential form:

1. Use log properties to combine into a single logarithm if needed.
2. Rewrite in exponential form: if $\log_b(\text{expression}) = c$, then $\text{expression} = b^c$.
3. Solve the resulting equation.
4. Check your answer. Plug it back in to make sure the argument of every logarithm is positive. Any solution that makes an argument zero or negative is extraneous and must be thrown out.

Example: Solve $\log_2(x + 3) = 5$.

1. Convert to exponential form: $x + 3 = 2^5 = 32$.
2. Solve: $x = 29$.
3. Check: $29 + 3 = 32 > 0$. Valid.

Applications of Exponential Models

Exponential functions model situations where a quantity multiplies by a constant factor over equal time intervals. The general form is:

$f(x) = a \cdot b^x$

• $a$ = initial value (the starting amount)
• $b$ = growth factor if $b > 1$, or decay factor if $0 < b < 1$
• $x$ = time or number of periods

Compound interest uses the formula:

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

Example: You invest $2,000 at 5% annual interest, compounded quarterly. How long until you have $3,000?

1. Set up: $3000 = 2000\left(1 + \frac{0.05}{4}\right)^{4t}$
2. Divide both sides by 2000: $1.5 = (1.0125)^{4t}$
3. Take $\ln$ of both sides: $\ln(1.5) = 4t \cdot \ln(1.0125)$
4. Solve: $t = \frac{\ln(1.5)}{4 \cdot \ln(1.0125)} \approx \frac{0.4055}{0.04969} \approx 8.16$ years

Exponential decay uses the formula:

$A(t) = A_0 \cdot e^{-kt}$

• $A(t)$ = amount remaining at time $t$
• $A_0$ = initial amount
• $k$ = decay constant (positive number)

This shows up in radioactive decay and similar problems where you solve for $t$ by isolating the exponential and then taking $\ln$ of both sides.

Characteristics of Exponential and Logarithmic Functions

Since these functions are inverses, their domains and ranges swap:

The domain restriction on logarithmic functions is why you always need to check solutions to logarithmic equations. If a solution makes the argument of any log zero or negative, it doesn't work.