Problem 15 - Applied Calculus

Question

For Problems \(1-16,\) solve for \(t\) using natural logarithms. $$7 \cdot 3^{t}=5 \cdot 2^{t}$$

Step-by-Step Solution

Answer

\(t = \frac{\ln(5/7)}{\ln(3) - \ln(2)}\)

Step 1: Isolate the exponential expressions

In the given equation, we have \(7 \cdot 3^t = 5 \cdot 2^t\). First, divide both sides by 7 to isolate the exponential expressions involving \(t\).\[3^t = \frac{5}{7} \cdot 2^t\]

Step 2: Take the natural logarithm of both sides

To eliminate the exponential expressions, take the natural logarithm (\(\ln\)) of both sides of the equation:\[\ln(3^t) = \ln\left(\frac{5}{7} \cdot 2^t\right)\]

Step 3: Apply the power rule for logarithms

Using the power rule for logarithms, \(\ln(a^b) = b \cdot \ln(a)\), we can simplify both sides:\[t \cdot \ln(3) = \ln\left(\frac{5}{7}\right) + t \cdot \ln(2)\]

Step 4: Isolate \(t\) on one side

Rearrange the equation to isolate \(t\). Move all terms involving \(t\) to one side:\[t \cdot \ln(3) - t \cdot \ln(2) = \ln\left(\frac{5}{7}\right)\]Factor \(t\) out of the left side:\[t(\ln(3) - \ln(2)) = \ln\left(\frac{5}{7}\right)\]

Step 5: Solve for \(t\)

Finally, solve for \(t\) by dividing both sides of the equation by \(\ln(3) - \ln(2)\):\[t = \frac{\ln\left(\frac{5}{7}\right)}{\ln(3) - \ln(2)}\]

Step 6: Simplify the expression if needed

The expression is simplified as is, but you can use a calculator to find the numerical approximation of \(t\) if needed.

Key Concepts

Exponential Functions

An exponential function is a mathematical expression in which a variable appears in the exponent. In simpler terms, it involves raising a constant base to the power of a variable. When dealing with exponential equations like the one in the exercise, recognizing this structure helps set the stage for solving it. For example, in the equation given, we have two different bases—3 and 2—both raised to the variable power of \( t \). The equation is: \[ 7 \cdot 3^t = 5 \cdot 2^t \]Exponential functions can be tricky because they grow very quickly. Some key properties of exponential functions include:

• The base of the exponent should be a positive constant not equal to 1.
• The variable appears only in the exponent.

These characteristics make exponential functions different from polynomial or linear functions. In math, understanding these properties is crucial in order to manipulate and solve such equations effectively.

Solve Equations

To solve equations involving exponential functions, we often use logarithms. The aim is to make the equation easier to handle and find the value of the unknown variable—in this case, \( t \). Rewriting the equation in simpler terms is the first step, as shown in the exercise. This often involves using basic algebra to isolate the term with the variable on one side. Here's the transformed equation:\[ 3^t = \frac{5}{7} \cdot 2^t \]This sets us up perfectly to apply logarithms, which are the inverse operations of exponentials.Applying logarithms (particularly natural logarithms, \( \ln \)) allows us to solve for \( t \). The fundamental idea here is that taking the logarithm of an exponential undoes the exponential, bringing the exponent down to the base-level:\[ \ln(3^t) = \ln\left(\frac{5}{7} \cdot 2^t\right) \]Breaking the equation in this manner allows us to move forward in solving for \( t \). This step-by-step simplification is critical when dealing with exponential equations.

Logarithmic Properties

Logarithmic properties are essential tools for solving exponential equations. When dealing with such equations, using the properties of logarithms helps simplify and solve them.Here are some key logarithmic properties that were used in the solution:

• Power Rule: \( \ln(a^b) = b \cdot \ln(a) \). This property allows us to bring down the exponent in front of the logarithm, simplifying the equation with respect to the variable \( t \).
• Product Rule: \( \ln(xy) = \ln(x) + \ln(y) \). Though not directly used in this exercise, understanding how to break down complex log expressions is useful.

In the exercise, the power rule was applied to get:\[ t \cdot \ln(3) = \ln\left(\frac{5}{7}\right) + t \cdot \ln(2) \]By applying these rules, we isolated the variable \( t \) to solve for it successfully. The idea is to use logarithms to transform multiplication or division in exponential equations into addition or subtraction, making it easier to isolate and solve for variables.