Problem 95 - Precalculus

Question

Solve for \(x\) in terms of \(b\) $$ \frac{1}{3} \log _{b}\left(x^{3}\right)+\frac{1}{2} \log _{b}\left(x^{2}-2 x+1\right)=2 $$

Step-by-Step Solution

Answer

The solution for \(x\) is \(x = \frac{1 + \sqrt{1 + 4b^2}}{2}\).

Step 1: Apply Logarithm Power Rule

We start by using the logarithm power rule, \(\log_b(a^c) = c \log_b(a)\). For the term \(\frac{1}{3} \log_b(x^3)\), we can simplify it to \(\log_b(x)\) because \(\frac{1}{3}\cdot 3 = 1\). For the term \(\frac{1}{2} \log_b(x^2 - 2x + 1)\), note that \(x^2 - 2x + 1\) can be rewritten as \((x-1)^2\). Applying the power rule again, \(\frac{1}{2}\cdot 2 \log_b(x-1) = \log_b(x-1)\). So, the equation becomes: \[ \log_b(x) + \log_b(x-1) = 2 \]

Step 2: Use Logarithm Product Rule

Now, we apply the logarithm product rule, \(\log_b(m) + \log_b(n) = \log_b(mn)\), to combine the logarithms on the left-hand side: \( \log_b(x) + \log_b(x-1) = \log_b(x(x-1))\). Therefore, the equation becomes: \[ \log_b(x(x-1)) = 2 \]

Step 3: Exponentiate to Eliminate the Logarithm

To eliminate the logarithm, exponentiate both sides of the equation. The base of the logarithm is \(b\), so we will have: \(x(x-1) = b^2\).

Step 4: Simplify and Solve the Quadratic Equation

Now, simplify \(x(x-1) = b^2\) to \(x^2 - x = b^2\). This is a quadratic equation in the form \(x^2 - x - b^2 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-1\), and \(c=-b^2\). Applying the quadratic formula gives: \[ x = \frac{1 \pm \sqrt{1 + 4b^2}}{2} \]

Step 5: Determine Valid Solutions

The problem requires solving for \(x\) in terms of \(b\). The quadratic formula results in two potential solutions: \(x = \frac{1 + \sqrt{1 + 4b^2}}{2}\) or \(x = \frac{1 - \sqrt{1 + 4b^2}}{2}\). However, \(\log_b(x)\) and \(\log_b(x-1)\) are only defined for positive values. Thus, \(x\) must be greater than 1 (since \(x-1\) must also be positive). Of the two solutions, only \(x = \frac{1 + \sqrt{1 + 4b^2}}{2}\) is valid, ensuring both terms inside the logarithms are positive.

Key Concepts

Logarithm Power Rule

The logarithm power rule is essential when dealing with equations involving exponents within logarithms. The rule states that \(\log_b(a^c) = c \log_b(a)\), allowing us to "move" the exponent to the front, simplifying complex expressions.

In the exercise, we see the application of this rule in two instances:

• For the term \(\frac{1}{3} \log_b(x^3)\), using the power rule results in \(\log_b(x)\), because multiplying \(\frac{1}{3}\cdot 3 = 1\).
• And, for \(\frac{1}{2} \log_b(x^2 - 2x + 1)\). Noticing that \(x^2 - 2x + 1\) is a perfect square, we write it as \((x-1)^2\). Applying the power rule here simplifies it to \(\log_b(x-1)\).

Thus, the initial complex logarithmic terms are greatly simplified. Mastering this rule not only makes complex equations manageable but also lessens computational errors.

Logarithm Product Rule

Another fundamental logarithmic property is the logarithm product rule, which combines two logarithms into one: \(\log_b(m) + \log_b(n) = \log_b(mn)\). This rule is quite handy when dealing with the sum of logarithms and makes equations simpler to solve.

In this exercise, after applying the power rule, we are left with:

• \(\log_b(x) + \log_b(x-1)\)

Using the product rule, this simplifies to \(\log_b(x(x-1))\). The beauty of this is it allows us to express the two original logarithms as a single one, making the next steps, such as exponentiating to remove the logarithm, much more straightforward.

Understanding and consistently using the product rule can turn seemingly challenging problems into more guided and logical processes, solidifying your problem-solving toolkit.

Quadratic Equations

Quadratic equations appear frequently in various mathematical contexts, and being adept at solving them is crucial. A typical quadratic equation takes the form \(ax^2 + bx + c = 0\). One common method for solving such equations is using the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In the exercise, after simplifying the logarithms, the equation becomes \(x^2 - x - b^2 = 0\). Here, \(a=1\), \(b=-1\), and \(c=-b^2\). Substituting these values into the quadratic formula gives two potential solutions. However, the context of logarithms requires we pick solutions ensuring positive values inside the original logarithmic expressions.

Therefore, understanding the nuances of quadratic equations will not only help in solving isolated quadratic problems but also in broader contexts, like when involving logarithms, ensuring valid solutions based on the problem’s constraints.