Problem 83 - College Algebra with Modeling and Visualization

Question

Solve each equation. Approximate answers to four decimal places when appropriate. $$5 \log _{7} 2 x=10$$

Step-by-Step Solution

Answer

The solution is \( x = 24.5 \).

Step 1: Isolate the Logarithmic Expression

First, divide both sides of the equation by 5 to isolate the logarithmic term. \[ \log_{7}(2x) = \frac{10}{5} \]Simplifies to:\[ \log_{7}(2x) = 2 \]

Step 2: Convert to Exponential Form

To remove the logarithm, convert the equation from logarithmic form to exponential form. From \(\log_{7}(2x) = 2\), convert to:\[ 2x = 7^2 \] This simplifies to:\[ 2x = 49 \]

Step 3: Solve for x

Now, solve for \(x\) by dividing both sides of the equation by 2.\[ x = \frac{49}{2} \]Simplifies to:\[ x = 24.5 \]

Key Concepts

Exponential Form

To solve logarithmic equations, converting them into exponential form can be very helpful. A logarithm essentially tells us the power to which a number (the base) must be raised to obtain another number. In the given problem, we started with the logarithmic equation \( \log_{7}(2x) = 2 \). Here, 7 is the base of the logarithm, 2 is the power (or exponent), and \(2x\) is the number obtained by raising 7 to that power.
To convert a logarithm into exponential form, rewrite it based on the definition of a logarithm. This means expressing it as \( \text{Base}^{\text{Exponent}} = \text{Number} \). Therefore, \( \log_{7}(2x) = 2 \) becomes \( 2x = 7^2 \). Converting into exponential form simplifies the problem, turning a logarithmic equation into a more manageable form. This is because exponentiation cancels out a logarithm.

Converting to exponential form is a critical step when solving logarithmic equations and it makes the next steps straightforward.

Solving Equations

Solving equations often requires isolating the variable you are trying to solve for. Once a logarithmic equation is transformed into exponential form, solving for the unknown variable becomes a simple matter of performing basic operations.
In our exercise, once we reach the equation \( 2x = 49 \), this can be solved by isolating \(x\). Here's how it's done:

• Divide both sides of the equation by the coefficient of \(x\) (which is 2 in this case). This gives us \( x = \frac{49}{2} \).
• Perform the division to find the value of \(x\). In our problem, this results in \( x = 24.5 \).

Breaking down the problem into small, manageable pieces helps us systematically find the solution. Once the exponential form is achieved, solving for the unknown becomes a simple task of algebraic manipulation, often involving basic operations like division. With practice, solving these types of equations can become easier and more intuitive.

Logarithms

Logarithms are the inverse of exponentiation, meaning they reverse the operation of raising a number to a power. Understanding logarithms is crucial because they help solve equations where the unknown variable is an exponent.
The notation \( \log_{b}(a) \) means "the power to which \(b\) must be raised to yield \(a\)." Logarithms provide a way to deal with large numbers by using the properties of exponents, such as the product, quotient, and power rules. These rules allow us to manipulate and solve complex equations that would be difficult otherwise.

When working with logarithms:

• Always remember the base: it tells you the number being repeatedly multiplied.
• The logarithmic equation can generally be simplified or solved by converting it into exponential form.
• For complex logarithmic expressions involving products, quotients, or powers, you can use logarithmic identities to simplify them first.

For most problems, especially in an educational context, logarithms should be approached step-by-step with careful attention to their definitions and properties.