Problem 56 - Algebra and Trigonometry

Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve: \(7^{x-1}=3 \cdot 2^{x+4}\)

Step-by-Step Solution

Answer

x = \frac{\ln(3) + 4 \ln(2) + \ln(7)}{\ln(7) - \ln(2)}

Step 1 - Rewrite the equation in logarithmic form

Given equation: \[ 7^{x-1} = 3 \times 2^{x+4} \]Take the natural logarithm (ln) of both sides:\[ \ln(7^{x-1}) = \ln(3 \times 2^{x+4}) \]

Step 2 - Simplify using logarithm properties

Use the properties of logarithms: \[ \ln(a^b) = b \ln(a) \text{ and } \ln(a \times b) = \ln(a) + \ln(b) \]Apply these properties to our equation:\[ (x-1) \ln(7) = \ln(3) + \ln(2^{x+4}) \]\[ (x-1) \ln(7) = \ln(3) + (x+4) \ln(2) \]

Step 3 - Distribute and combine like terms

Distribute \(\ln(7)\) and \(\ln(2)\):\[ x \ln(7) - \ln(7) = \ln(3) + x \ln(2) + 4 \ln(2) \]Move all terms involving \(x\) to one side:\[ x \ln(7) - x \ln(2) = \ln(3) + 4 \ln(2) + \ln(7) \]

Step 4 - Factor out x and solve for x

Factor out \(x\) on the left:\[ x (\ln(7) - \ln(2)) = \ln(3) + 4 \ln(2) + \ln(7) \]Solve for \(x\):\[ x = \frac{\ln(3) + 4 \ln(2) + \ln(7)}{\ln(7) - \ln(2)} \]

Key Concepts

Properties of Logarithms

Logarithms are highly useful in mathematics, especially for simplifying complex expressions and solving equations. Understanding their properties can make many math problems easier.
Here are some key properties you should know:

• Product Property: This property states that the logarithm of a product is the sum of the logarithms of its factors:\[ \log_b(MN) = \log_b(M) + \log_b(N) \]
• Quotient Property: This states that the logarithm of a quotient is the difference between the logarithms of the numerator and denominator:\[ \log_b \left( \frac{M}{N} \right) = \log_b(M) - \log_b(N) \]
• Power Property: This shows that the logarithm of a number raised to a power is the power times the logarithm of the number:\[ \log_b(M^k) = k \log_b(M) \]

In the given equation, we used these properties to simplify and solve it:

• Product Property: Applied to \( \ln(3 \cdot 2^{x+4}) \)
• Power Property: Applied to \( \ln(7^{x-1}) \) and \( \ln(2^{x+4}) \)

Solving Exponential Equations

Exponential equations can look intimidating, but they become manageable once you know the steps to solve them. Here are the steps we followed:
First, identify the exponential equation you need to solve. For instance, in this problem: \( 7^{x-1} = 3 \cdot 2^{x+4} \).

Next, to simplify such equations, take the natural logarithm (ln) of both sides. This makes it easier to handle because the logarithm allows us to bring down the exponents:
\[ \ln(7^{x-1}) = \ln(3 \cdot 2^{x+4}) \]
Use the properties of logarithms mentioned earlier to simplify the equation further:
\[ (x-1) \ln(7) = \ln(3) + (x+4) \ln(2) \]
Combine like terms and isolate the term with the variable (x) on one side. Factor out x, if necessary:
\[ x (\ln(7) - \ln(2)) = \ln(3) + 4 \ln(2) + \ln(7) \]
Finally, solve for x by dividing both sides by the coefficient that multiplies x:
\[ x = \frac{\ln(3) + 4 \ln(2) + \ln(7)}{\ln(7) - \ln(2)} \].
These steps will help solve exponential equations efficiently.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a special kind of logarithm with the base \( e \). The constant \( e \approx 2.71828 \) is an important number in mathematics, similar to \( \pi \). The natural logarithm has several useful properties:

• Logarithm of 1: \( \ln(1) = 0 \)
  
• Natural exponent base: \( \ln(e) = 1 \)
  
• Like other logarithms, it follows properties such as the product, quotient, and power rules:

The natural logarithm is particularly handy for dealing with exponential equations, especially when the base of the exponent is \( e \).
In our problem, we took the natural logarithm on both sides of the equation: \( \ln(7^{x-1}) = \ln(3 \cdot 2^{x+4}) \). This step transformed the exponential forms into linear terms, which are much easier to manipulate and solve.
Being comfortable with the natural logarithm and its properties can make a wide range of mathematical problems easier to tackle.