Problem 51 - Algebra 2 and Trigonometry

Question

Solve each equation for the variable. \(\log _{5} x+\log _{5} x=\log _{5} 625\)

Step-by-Step Solution

Answer

The solution is \(x = 25\).

Step 1: Combine Logarithms on the Left Side

The given equation is \(\log_{5} x + \log_{5} x = \log_{5} 625\). Combine the logarithms on the left-hand side using the property: \(\log_{b}a + \log_{b}c = \log_{b}(ac)\). This gives us \(\log_{5}(x^2) = \log_{5} 625\).

Step 2: Set the Arguments Equal

Since the bases of the logarithms on both sides of the equation are the same, we can drop the logarithms and set their arguments equal to each other: \(x^2 = 625\).

Step 3: Solve for the Variable

To solve \(x^2 = 625\), take the square root on both sides: \(x = \pm 25\). Since \(x\) represents a value inside a logarithm, we only consider the positive solution: \(x = 25\).

Key Concepts

Properties of Logarithms

Logarithms are powerful mathematical tools used to simplify complex multiplicative operations into addition and subtraction. Several fundamental properties characterize them, and these properties can help us solve logarithmic equations.

Some key properties are:

• Product Property: \(\log_{b}(a) + \log_{b}(c) = \log_{b}(a \times c)\). This property was used in the exercise to combine the two logarithms: \(\log_{5} x + \log_{5} x = \log_{5}(x^2)\).
• Quotient Property: \(\log_{b}(a) - \log_{b}(c) = \log_{b}(\frac{a}{c})\).
• Power Property: \(\log_{b}(a^c) = c \cdot \log_{b}(a)\).

These properties are useful in rewriting logarithmic expressions, resulting in much simpler forms to work with. Whenever you see logarithms, think about these properties first—they are your shortcuts to simplifying and solving problems.

Solving Equations

When it comes to solving logarithmic equations, the goal is to isolate the variable. This means representing your equation in a form that makes it easier to pinpoint the variable's value.

Start by using the properties of logarithms to simplify any multi-logarithmic terms. In exercises like the one provided, you combine terms to have a single logarithm on each side of the equation. This allows you to set the arguments equal, once like-bases logarithms are involved on both sides. Remember, you can remove the logarithms when the bases are the same on both sides.

After you isolate the variable, make sure to check if the initial conditions (like the base of the logarithm) allow the solution to be valid. It's important to remember that the argument of a logarithm must be positive, so certain solutions might need exclusion if they don't meet this criterion. This step is crucial to confirm that your final answer is valid.

Combining Logarithms

Combining logarithms simplifies complex-looking equations into manageable ones. In the exercise solution, the key task was to combine \(\log_{5} x + \log_{5} x\). Using the properties of logarithms, specifically the Product Property, we combined them into \(\log_{5}(x^2)\).

By combining the logarithms, we reduce the number of terms and simplify both the calculation and the equation-structuring.

• This operation makes it easier to equate and isolate variables.
• It often turns longer expressions into single, more concise terms.

Combining logarithms is essential because it helps to manage equations before solving them, often transforming the equation into a more recognizable or traditional form.

Square Root

Understanding the square root operation is crucial in solving the equation once you've simplified it. In our exercise, we arrived at \(x^2 = 625\) after using the properties of logarithms and solving strategies.

To solve this, we employ the square root operation. Taking the square root of both sides yields two answers: \(x = 25\) and \(x = -25\). However, since the context of logarithms requires the input \(x\) to be positive, we only consider \(x = 25\).

Keep in mind that the square root function converts a squared number back to its original value (or its positive counterpart, in most mathematical contexts). It's an essential technique not just in solving quadratic equations but also in other algebraic manipulations including those in geometry and calculus. Knowing when and how to apply this operation is key to successfully working through many types of equations.