Problem 8 - College Algebra and Calculus: An Applied Approach

Question

Solve for \(x\).\(\log _{5} 5 x=2\)

Step-by-Step Solution

Answer

The solution for \(x\) is \(5\).

Step 1: Rewrite the logarithm as an exponent

By the definition of logarithms, \(\log _{5}(5x) = 2\) implies that \(5^2 = 5x\).

Step 2: Simplify the equation

Calculating \(5^2\) yields \(25 = 5x\).

Step 3: Divide by 5

Finally, to isolate \(x\), divide both sides of the equation by \(5\). So, \(x = 25/5 = 5\).

Key Concepts

Logarithms

Logarithms are the inverse operations of exponents, much like subtraction is the inverse of addition. When we take the logarithm of a number, we are essentially asking the question: "To what power must we raise a specific base to obtain this number?" For example, in the expression \(\log_{5}(5x) = 2\), the base is 5, and the logical question is "5 raised to what power equals \(5x\)?" When we solve this, we imply that raising 5 to the power of 2 yields \(5x\).

• Base: The number that is repeated in multiplication (5 in our example).
• Argument: The number whose logarithm is being calculated (\(5x\) here).
• Logarithm (result): The power to which the base must be raised (which is 2).

Understanding this concept is crucial because it transforms complex multiplication problems into simpler addition tasks within logarithms.

Exponents

Exponents are powerful tools that let us express repeated multiplication compactly. In mathematical terms, when we say \(5^2\), we mean 5 multiplied by itself, which equals 25. Exponents provide a convenient way to handle large calculations by allowing a single number to represent multiple factors.

The relationship between logarithms and exponents is crucial in algebra. As logarithms are inverse operations of exponents, understanding one can help solve problems involving the other. In the problem \(\log_{5}(5x) = 2\), when we translate it to an exponential form, it becomes \(5^2 = 5x\). This transition shows how understanding the interplay between these two concepts can simplify solving equations. Always remember that the exponent indicates the number of times the base near it is multiplied by itself.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to find an unknown value, in this case, \(x\). It is about using various algebraic properties to isolate the variable on one side of the equation. In our solution \(5^2 = 5x\), we start by solving for \(5^2\), which is 25.

The next step is crucial: isolating \(x\). Divide both sides of the equation by the coefficient of \(x\), which in our problem is 5, giving us \(x = \frac{25}{5}\). Simplifying this division, we get \(x = 5\). This is a clear example of algebraic manipulation at work, where solving the equation requires systematic steps:

• Performing operations equally on both sides.
• Simplifying to make the solution straightforward.
• Applying inverse operations to isolate the variable.

Each step simplifies the problem closer to the solution for \(x\).