Question

The number of solutions of \(\log _{9}(2 x-5)=\log _{3}(x-4)\) is: (a) 0 (b) 1 (c) 2 (d) 3

Short answer

a) 0 b) 1 c) 2 Answer: (b) 1

Step-by-step solution

Rewrite the logs with the same base

We can rewrite the logs with the same base to make them easier to compare. Since the base 9 can be expressed as 3 squared, we can rewrite the equation as: $$\log _{3^2}(2 x-5)=\log _{3}(x-4)$$ Now, using the property \(\log _{a^b}(x)= \frac{1}{b}\log _{a}(x)\), the equation becomes: $$\frac{1}{2} \log _{3}(2 x-5)=\log _{3}(x-4)$$

Remove the logs by exponentiating both sides

To remove the logs, we can exponentiate both sides of the equation with base 3: $$3^{\frac{1}{2} \log _{3}(2 x-5)}=3^{\log _{3}(x-4)}$$ From the property \(a^{\log _{a}(x)}=x\), we obtain: $$\sqrt{2x-5}=x-4$$

Solve the equation for x

To solve the equation, first square both sides to get rid of the square root: $$(2x-5)=(x-4)^2$$ Now, we will expand the square and simplify the equation: $$2x-5=x^2-8x+16$$ Rearrange the equation to obtain a quadratic equation in the form of \(ax^2+bx+c=0\): $$x^2-10x+21=0$$

Find the solutions for x

Now, we need to find the solutions for x. In order to do that, we can factor the above quadratic equation: $$(x-7)(x-3)=0$$ So, we have two possible solutions for x: 1) \(x=7\) 2) \(x=3\)

Verify the solutions are valid

Finally, we need to check if the solutions we found are valid by substituting them back into the original equation: For \(x=7\): $$\log _{9}(2 (7)-5)=\log _{3}((7)-4)$$ $$\log _{9}(9)=\log _{3}(3)$$ This solution is valid. For \(x=3\): $$\log _{9}(2 (3)-5)=\log _{3}((3)-4)$$ $$\log _{9}(1)=\log _{3}(-1)$$ This solution is not valid because the logarithm of a negative number is undefined. Thus, there is only one valid solution for the given equation. The answer is (b) 1.

Key concepts

Base Transformation

When working with logarithmic equations, transforming the base of the logarithms to the same number can simplify the problem. Often, using the property \( \log_{a^b}(x) = \frac{1}{b}\log_a(x) \) helps.

In the original exercise, we transformed a logarithm with base 9 into base 3, since 9 is the square of 3.
This makes it easier to handle both sides of the equation simultaneously and compare them directly.

Here's a quick reminder on base transformation steps:

• Identify a common base for both logarithms.
• Express the bases as powers of the common base, if possible.
• Utilize the formula \( \log_{a^b}(x) = \frac{1}{b}\log_a(x) \) to rewrite the equation.

Transforming the base not only helps in simplifying the problem but also in finding accurate solutions.

Quadratic Equations

After simplifying the original logarithmic equation, we often end up with a quadratic equation.

Quadratic equations typically take the form \( ax^2 + bx + c = 0 \). In this exercise, after shifting terms, we arrived at the equation \( x^2 - 10x + 21 = 0 \).

Here's how to tackle quadratic equations:

• Try factoring the quadratic into two binomials: \( (x - m)(x - n) = 0 \).
• Identify values of \( x \) that satisfy each binomial equation, giving potential solutions.

In this example, the factors turned out to be \( (x-7)(x-3) \), leading to solutions \( x = 7 \) and \( x = 3 \). Always remember, not all solutions are valid without checking back with the original context.

Properties of Logarithms

Understanding properties of logarithms is crucial when solving equations involving logarithmic functions.

Some key properties include:

• \( \log_a(x) = y \), means \( a^y = x \).
• \( a^{\log_a(x)} = x \).
• Logarithms are undefined for zero or negative numbers.

In this exercise, a property used was \( a^{\log_a(x)} = x \), which allowed us to remove the logarithm entirely and work with a simpler equation.

Moreover, remember that verifying each solution is essential since it's possible the transformed equations produce extraneous solutions, especially if the domain of the original equation restricts specific values. Always verify substituting back into the original equation to ensure a solution is valid.