Problem 55 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

\(\log _{4}(x-1)=\log _{2}(x-3)\)

Step-by-Step Solution

Answer

The solution to the equation is \(x = 9\).

Step 1: Rewrite using the change of base formula

Using the change of base formula, \(\log _{a}b = \frac{\log b}{\log a}\), we could rewrite both sides of the equation as: \(\frac{\log (x-1)}{\log 4} = \frac{\log (x-3)}{\log 2}\)

Step 2: Simplify the equation

\(\log 4 = 2\) given how \(\log 2^2 = 2\). The formula can be then written as: \(\frac{\log (x-1)}{2} = \log (x-3)\)

Step 3: Multiply both sides by 2 and solve for x

In multiplying the equation by 2, we would have: \(\log (x-1) = 2\log (x-3)\), which simplifies to: \(\log (x-1) = \log [(x-3)^2]\). By setting \(x-1 = (x-3)^2\), we obtain \(x^2-8x+9 = 0\), a quadratic equation which we can solve for real number roots.

Step 4: Find the roots of the equation

By factoring the equation, we have \((x-1)(x-9) = 0\). Therefore, the roots are \(1\) and \(9\). However, we need to check if these roots are valid by substituting these values back into the original equation and checking whether both sides are still equal.

Step 5: Check the solutions

Let's substitute \(x = 1\) into the original equation: \(\log _{4}(1-1) =? \log _{2}(1-3)\). The left-hand side is undefined, therefore \(x = 1\) is not a solution. Now, if we substitute \(x = 9\) into the original equation, we have: \(\log _{4}(9-1) = \log _{2}(9-3)\), which simplifies to \(\log _{4}(8) = \log _{2}(6)\) and is true. Thus, \(x = 9\) is the only solution.

Key Concepts

Change of Base Formula

The change of base formula is a useful tool in mathematics when working with logarithms. It allows us to rewrite a logarithm in terms of any other base. This is particularly helpful when the calculator you are using does not support every base.
Let's see how it works. If you have a logarithm expressed as \(\log_a b\), the change of base formula tells us:

• \(\log_a b = \frac{\log_c b}{\log_c a}\)

Here, you can choose your base \(c\) freely, but common choices are 10 and \(e\) (for the natural logarithm), because these are available on most calculators.
In our original exercise, using this formula, we transformed the left and right sides of the equation. We went from \(\log_4(x-1) = \log_2(x-3)\) to an equation using base 10:

• \(\frac{\log(x-1)}{\log 4} = \frac{\log(x-3)}{\log 2}\)

Using the change of base formula can thus simplify computations and resolve log equations by using a standard base available to us.

Solving Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). To solve them, you can use various methods like factoring, completing the square, or the quadratic formula.
Let's delve into these a little more:

• Factoring: If the quadratic can be factorized easily into two binomials, it's often the quickest way. For example, \(x^2 - 8x + 9 = 0\) factors into \((x-1)(x-9) = 0\).
• Quadratic Formula: When factoring is not straightforward, you can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).

In our step-by-step solution, the equation \(x-1 = (x-3)^2\) was simplified to \(x^2 - 8x + 9 = 0\), which was easy to factor. This gave us potential solutions \(x = 1\) and \(x = 9\). Always remember to check these solutions in the original equation because quadratic approximations can introduce extraneous roots.

Logarithmic Properties

Logarithmic properties are rules that simplify expressions and solve logarithmic equations. Understanding these properties allows you to manipulate logarithmic expressions in algebraic operations.
Here are some essential properties:

• Product Rule: \(\log_b(MN) = \log_b M + \log_b N\)
• Quotient Rule: \(\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N\)
• Power Rule: \(\log_b(M^n) = n \cdot \log_b M\)

In our solved problem, we used the properties of logarithms to simplify \(\log (x-1) = 2\log (x-3)\) by using the power rule to transform it to \(\log (x-1) = \log [(x-3)^2]\).
This simplification enabled us to equate the arguments, leading to a quadratic equation that we could solve to find \(x\). Mastery of these logarithmic properties extends your ability to solve complex equations with ease.