Problem 16 - Precalculus Mathematics for Calculus

Question

Find the solution of the exponential equation, rounded to four decimal places. $$3^{x / 14}=0.1$$

Step-by-Step Solution

Answer

\(x \approx -29.3426\)

Step 1: Understand the Exponential Equation

The given exponential equation is \(3^{x/14} = 0.1\). Our task is to find the value of \(x\) that satisfies this equation.

Step 2: Apply the Logarithm to Both Sides

To solve for \(x\), we need to eliminate the exponent. We do this by taking the logarithm of both sides of the equation. This will help us bring down the exponent for easier manipulation.\[\log(3^{x/14}) = \log(0.1)\]

Step 3: Use the Property of Logarithms

Using the logarithmic identity \(\log(a^b) = b\log(a)\), we can rewrite the left side of the equation.\[\frac{x}{14}\cdot \log(3) = \log(0.1)\]

Step 4: Solve for x

Now, isolate \(x\) by first multiplying both sides by 14, and then divide by \(\log(3)\).\[x = 14 \cdot \frac{\log(0.1)}{\log(3)}\]

Step 5: Calculate the Value Using Logarithm Properties

Perform the calculation to find \(x\). Using the property \(\log(0.1) = \log(10^{-1}) = -1\), substitute and compute:\[x = 14 \cdot \frac{-1}{\log(3)}\]

Step 6: Compute the Numerical Value

Using a calculator to find \(\log(3)\), approximately 0.4771:\[x = 14 \cdot \frac{-1}{0.4771} \approx -29.3426\]

Step 7: Round the Result

Round the calculated value of \(x\) to four decimal places to get \[x \approx -29.3426\].

Key Concepts

Logarithms

Logarithms play a crucial role in solving many mathematical equations, especially when dealing with exponential functions. They were invented in the 17th century to simplify calculations by transforming complex multiplications into simpler additions. In simple terms, a logarithm answers the question: to what power must a specific base be raised, in order to obtain a particular number?

To understand logarithmic functions, let's consider the expression \( \log_{b}(a) \), which asks what power \( b \) must be raised to, to get \( a \). Here, \( b \) is often referred to as the "base." For example, \( \log_{10}(100) = 2 \) because \( 10^2 = 100 \). The base 10 logarithm is often used in calculations, and when no base is written, it is generally assumed to be 10.

Logarithms are particularly helpful in solving exponential equations because they provide a method to "bring down" the exponent, making the equation more manageable to solve. As in our exercise, applying logs made it possible to isolate the variable \( x \) and solve the equation.

Properties of Logarithms

There are several important properties of logarithms that simplify calculations and make manipulation of logarithmic terms straightforward. These properties allow for exponent rules to be applied in reverse, which is extremely helpful in the process of solving equations.

Here are some key properties to remember:

• Product Rule: \( \log_b(MN) = \log_b(M) + \log_b(N) \). This helps when multiplying likes of expressions, turning them into sum of logs.
• Quotient Rule: \( \log_b(M/N) = \log_b(M) - \log_b(N) \). This is useful when dividing expressions, converting them into differences of logs.
• Power Rule: \( \log_b(M^n) = n \cdot \log_b(M) \). By using this rule, we can bring down the exponent \( n \) as a multiplier, simplifying the equation.

Notice how we used the power rule in our exercise: \( \log(3^{x/14}) = \frac{x}{14} \cdot \log(3) \). This transformation allows us to handle the equation without the exponent looming over as a complex hurdle to be overcome.

Solving Exponential Equations

Exponential equations can be intimidating at first glance because they involve variables in the exponent, but with logarithms, these problems become much more approachable. Here’s a step-by-step approach:

• Identify the expression in the equation. In our example, it was \( 3^{x/14} = 0.1 \).
• Apply the logarithm to both sides of the equation. This will allow the exponents to be brought down and the equation transformed. We used \( \log(3^{x/14}) = \log(0.1) \).
• Use the properties of logarithms, like the power rule, to simplify the expression further and isolate the variable. For instance, \( \frac{x}{14} \cdot \log(3) = \log(0.1) \).
• Finally, solve for the variable \( x \) by isolating it on one side. Perform any calculations necessary, such as multiplying, dividing, or using a calculator for the logarithmic values, and solve for the unknown \( x \). In this case, it yielded \( x \approx -29.3426 \).

By following these steps, the intimidating form of an exponential equation becomes accessible and straightforward, making them much easier to work with. Always remember to round if the problem demands a precise numerical result.