Problem 12 - Algebra and Trigonometry

Question

Find the solution of the exponential equation, correct to four decimal places. $$ 2^{3 x}=34 $$

Step-by-Step Solution

Answer

The solution is approximately \(x = 1.6963\).

Step 1: Understand the Equation

The equation given is an exponential equation of the form \(2^{3x} = 34\). Our goal is to solve for \(x\).

Step 2: Apply Logarithm to Both Sides

To solve for \(x\), we'll apply the logarithm to both sides of the equation. Using logarithm base 10 (common logarithm), this becomes \(\log(2^{3x}) = \log(34)\).

Step 3: Use Logarithmic Identity

Use the identity \(\log(a^b) = b \cdot \log(a)\) to simplify the left side: \(3x \cdot \log(2) = \log(34)\).

Step 4: Solve for x

Isolate \(x\) by dividing both sides of the equation by \(3 \cdot \log(2)\). So, \(x = \frac{\log(34)}{3 \cdot \log(2)}\).

Step 5: Calculate the Value

Calculate the values using a calculator. \(\log(34) \approx 1.5315\) and \(\log(2) \approx 0.3010\). Substitute these values into the equation: \(x = \frac{1.5315}{3 \cdot 0.3010}\).

Step 6: Simplify the Expression

Simplify the expression \(x = \frac{1.5315}{0.903}\) to find \(x \approx 1.6963\).

Key Concepts

Logarithmic Functions

Logarithmic functions are mathematical functions that help us deal with exponential relationships. They are particularly useful when working with exponential equations. The basic idea is that a logarithm answers the question: to what exponent do we need to raise a certain base to get a particular number?
For example, the logarithm base 10 of 100 is 2, because 10 raised to the power of 2 equals 100. This can be expressed as \( \log_{10}(100) = 2 \).
Logarithmic functions have the property of converting multiplication into addition and exponents into multiplication, which simplifies many types of calculations. This is especially useful when you're trying to solve equations that involve exponentials, as the logarithm function can provide a straightforward way to "unravel" the powers and make the solved variable more apparent.
When choosing a logarithmic base, base 10 and the natural base, e (Euler's number), are the most common choices. Each has its own applications, but using base 10 (common logarithm) simplifies many expressions, as calculators often provide direct functions for them.

Solving Equations

Solving equations is a core mathematical skill that involves finding the value of a variable that makes the equation true. When it comes to exponential equations, such as \(2^{3x} = 34\), we need specific strategies because the variable is inside the exponent.
Applying logarithms is a powerful technique in this context, as it allows us to bring the exponent down and convert the equation into a linear form. This conversion into linear terms makes it easier to isolate the variable and thus solve the equation.
Consider the exponential equation \(2^{3x} = 34\). By applying the logarithm, the equation changes significantly. This step, \(\log(2^{3x}) = \log(34)\), relies on logarithmic identities to bring down the exponent. Once you have a linear equation, you can use algebra to solve for the variable in question.
Thus, the conversion and application of inverse operations are key in solving such equations.

Using Logarithms

Using logarithms is not only essential for dealing with exponential equations but also helps in various scientific and engineering applications. In solving equations, we leverage properties of logarithms to simplify and solve for the variable.
The step where we use the identity \(\log(a^b) = b \cdot \log(a)\) is crucial. This allows us to take an exponent and transform it into a multiply operation, making it easier to manipulate algebraically. In our example, this transforms \(\log(2^{3x})\) into \(3x \cdot \log(2)\).
Next, solving for the variable involves isolating it. We do this by dividing both sides by the coefficient term, like \(3 \cdot \log(2)\), resulting in simpler direct computations like \(x = \frac{\log(34)}{3 \cdot \log(2)}\).
Finally, accurate computation of logarithm values, often done with a calculator, is the last step. By substituting these values and simplifying, we can solve the equation successfully. Thus, understanding how to use logarithms in this structured way enables successful problem-solving in a wide range of logarithmic contexts.