Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use natural logarithms to solve each of the exponential equations. Hint: To solve \(3^{x}=11\), take \(\ln\) of both sides, obtaining \(x \ln 3=\ln 11 ;\) then \(x=(\ln 11) /(\ln 3) \approx 2.1827 .\) $$ 5^{2 s-3}=4 $$

Step-by-Step Solution

Answer

The approximate value of \(s\) is 2.216.

Step 1: Write the Equation

We start with the given exponential equation: \[ 5^{2s-3} = 4 \]

Step 2: Apply Natural Logarithm

Take the natural logarithm on both sides of the equation to bring down the exponent:\[ \ln(5^{2s-3}) = \ln(4) \]

Step 3: Use Logarithm Property

Use the property of logarithms \(\ln(a^b) = b\ln(a)\) to simplify:\[ (2s-3) \ln(5) = \ln(4) \]

Step 4: Isolate the Term with Variable

Divide both sides by \(\ln(5)\) to isolate the term with the variable:\[ 2s - 3 = \frac{\ln(4)}{\ln(5)} \]

Step 5: Solve for the Variable

Add 3 to both sides and divide by 2 to solve for \(s\):\[ 2s = \frac{\ln(4)}{\ln(5)} + 3 \]\[ s = \frac{1}{2} \left( \frac{\ln(4)}{\ln(5)} + 3 \right) \]

Step 6: Calculate the Approximate Value

Compute the value using a calculator:\[ s \approx \frac{1}{2} \left( \frac{1.3863}{1.6094} + 3 \right) \approx 2.216 \]

Key Concepts

Exponential Equations

Exponential equations are those where the unknown variable appears in the exponent. They are common in many applications like compound interest, population growth, and radioactive decay. The general form of an exponential equation can be shown as \(a^{x} = b\). These equations can be challenging to solve directly due to the variable's placement in the exponent. In these cases, logarithms become a useful tool as they allow us to bring the variable down to a level where algebraic techniques can be applied. To solve exponential equations, we use properties of logarithms, especially working with natural logarithms, which are logs with base \(e\).

Logarithm Properties

Logarithms are the inverse of exponential functions and possess several properties that make them powerful tools for solving equations. The essential properties of logarithms that aid in solving exponential equations include:

• Product Property: \(\ln(ab) = \ln(a) + \ln(b)\)
• Quotient Property: \(\ln \left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
• Power Property: \(\ln(a^b) = b \cdot \ln(a)\)

For solving exponential equations, the power property is most often used. It allows one to transform an equation in which the variable is in the exponent into one where the variable is a coefficient. This transformation makes it possible to use algebra to isolate the variable and solve the equation.

Solving Equations with Logarithms

When solving exponential equations using logarithms, the first step is typically to apply the natural logarithm (\(\ln\)) to both sides of the equation. For example, in the equation \(5^{2s-3} = 4\), applying \(\ln\) gives \(\ln(5^{2s-3}) = \ln(4)\). Using the power property, we can simplify the left side to \((2s-3) \ln(5)\). From this point, our variable \(s\) is no longer in an exponent. The remaining steps involve isolating \(s\), often by dividing or adding constants to both sides of the equation. Finally, solving the simplified algebraic equation gives us the exact solution, which can often be expressed as a logarithm of a ratio, such as \(s = \frac{1}{2} \left(\frac{\ln(4)}{\ln(5)} + 3\right)\).

Approximate Solutions

Exact solutions involving logarithms can sometimes be cumbersome to use in practical applications. Thus, finding an approximate solution is often useful. Approximations require the use of a calculator to evaluate the natural logarithms numerically. Using our example, \(\ln(4)\) and \(\ln(5)\) can be estimated as approximately 1.3863 and 1.6094, respectively. These values are then substituted back into our algebraic expression for \(s\), resulting in an approximate value: \(s \approx \frac{1}{2} \left(\frac{1.3863}{1.6094} + 3\right)\). So, \(s \approx 2.216\). Approximate solutions are helpful when a numerical answer is required and they provide insight into the practicality of exact algebraic solutions.