Problem 62 - Precalculus with Limits

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. \(\log 8 x-\log (1+\sqrt{x})=2\)

Step-by-Step Solution

Answer

The approximate solution to three decimal places is \(x = 176.287\)

Step 1: Apply the rule of logarithms

The rule of logs states that the difference of two logs translates to the division of their arguments. So, we transform the equation \(\log 8 x - \log (1+\sqrt{x})=2\) into \(\log \(\frac{8x}{1 + \sqrt{x}}\)=2\)

Step 2: Remove the logarithm

The logarithm base is 10. Applying the principles of logarithms, we can remove the log from the equation which translates to \(\frac{8x}{1 + \sqrt{x}} = 10^2\), resulting into \(\frac{8x}{1 + \sqrt{x}} = 100\)

Step 3: Solve for x

Cross multiply and rearrange the equation to solve for x as follows: \(8x = 100 * (1 + \sqrt{x})\) → \(8x = 100 + 100\sqrt{x}\) → \(0 = 100 - 8x + 100\sqrt{x}\). Substitute \(a = \sqrt{x}\), Therefore, \(0 = 100 - 8a^2 + 200a\), rearranging gives quadratic equation \(0 = 8a^2 - 200a + 100\), dividing through by 8 gives \(a^2 - 25a + 12.5 = 0\). By quadratic formula, \(a = \frac{-(-25) ± \sqrt{(-25)^2 - 4(12.5)}}{2}\) \(a = \frac{25 ± \sqrt{625 - 50}}{2}\) = \(a_{1,2} = 13.263, 0.947\). Remember that \(a = sqrt(x)\), thus \(x_{1,2} = 176.287, 0.897\). The solutions for the equation are \(x = 176.287\) and \(x = 0.897\). But as the logarithm of zero is undefined, the only valid solution is \(x = 176.287\).

Key Concepts

Logarithm Properties

Logarithms are incredibly useful in mathematics, especially when dealing with products, quotients, and powers. One fundamental property is that the logarithm of a quotient is equal to the difference of the logarithms. In our exercise, this property allows us to rewrite \(\log 8x - \log (1 + \sqrt{x}) = 2\) as \(\log \left(\frac{8x}{1 + \sqrt{x}}\right) = 2\), simplifying the problem significantly.

This transformation is based on the rule \(\log_b(\frac{M}{N}) = \log_b(M) - \log_b(N)\), where \(b\) is the base of the logarithm, \(M\) and \(N\) are any positive real numbers. Other properties include \(\log_b(MN) = \log_b(M) + \log_b(N)\) for products, and \(\log_b(M^k) = k\cdot\log_b(M)\) for powers. These properties often come into play when solving logarithmic equations and should be remembered as they make the process much easier.

Algebraic Methods

When faced with logarithmic equations, algebraic methods become the cornerstone of finding solutions. In our example, once the logarithm is isolated, we use the principle of logarithms to convert the equation into a more familiar algebraic form. By setting \(\frac{8x}{1 + \sqrt{x}} = 10^2\), we remove the logarithmic operation and proceed with a cross-multiplication that simplifies the equation into a form where \(x\) can be isolated.

Another algebraic method used here is substitution, which is particularly helpful when dealing with radicals or higher powers of a variable. In this case, we substitute \(a = \sqrt{x}\) in order to transform the equation into a quadratic form. Substitution makes the equation more manageable and brings us closer toward finding the value of \(x\). Always revert the substitution at the end to return to the original variable.

Quadratic Equations

Quadratic equations, which are of the form \(ax^2 + bx + c = 0\), appear often in various areas of mathematics. In our exercise, we arrive at a quadratic equation after substituting \(a = \sqrt{x}\) into our earlier expression. The standard method of solving a quadratic equation involves factoring, completing the square, or using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

In the given example, the quadratic formula is used to find the solutions for \(a\). These values of \(a\) are then squared to get the solutions for \(x\). This exercise demonstrates that even though we started with a logarithmic equation, understanding how to solve quadratic equations is essential since many algebraic problems eventually lead to this form. Furthermore, it's important to check each solution in the original equation to ensure they are valid, as some solutions may not satisfy the original logarithmic conditions.