Logarithmic Functions Properties - 1

Question 1: The Power Rule of Logarithms

The Question: For $0 < a < 1$ and $M > 0$, $\log_a(M^2) = ?$

• 1
• $2\log_a M$
• $2\log_M a$
• $\log_a 3M$

Core Concept: This is a direct application of the Power Rule: $\log_b(M^p) = p \cdot \log_b M$.

Detailed Solution:

1. The expression is $\log_a(M^2)$.
2. In this expression, the base is $a$, the argument is $M$, and the power is $2$.
3. According to the Power Rule, we can bring the exponent (2) to the front of the logarithm as a multiplier.
4. This gives the result: $2 \log_a M$.

Final Answer: $2\log_a M$

Question 2: Solving a Logarithmic Equation

The Question: Solve for $x$: $\log_{12}(x-2) + \log_{12}(x+2) = 1$

• $x = -4$
• $x = 4$
• $x = 2$
• $x = -2$

Detailed Solution:

1. Determine the Domain: Before solving, we find the values of $x$ for which the equation is defined.

   • For $\log_{12}(x-2)$, the argument must be positive: $x-2 > 0 \implies x > 2$.
   • For $\log_{12}(x+2)$, the argument must be positive: $x+2 > 0 \implies x > -2$.
   • To satisfy both conditions, we must have $x > 2$. Any solution we find must meet this condition.

2. Combine the Logarithms: The left side is a sum of two logs with the same base. We can use the Product Rule to combine them.

   • $\log_{12}((x-2)(x+2)) = 1$
   • $\log_{12}(x^2 - 4) = 1$

3. Convert to Exponential Form: The definition of a logarithm is $y = \log_b x \iff b^y = x$.

   • In our case, the base is 12, the exponent is 1, and the argument is $x^2-4$.
   • $12^1 = x^2 - 4$

4. Solve for x:

   • $12 = x^2 - 4$
   • $16 = x^2$
   • $x = \pm\sqrt{16} \implies x=4$ or $x=-4$.

5. Check for Extraneous Solutions: We must check our solutions against the domain requirement ($x > 2$).

   • $x=4$: This is valid because $4 > 2$.
   • $x=-4$: This is not valid because $-4$ is not greater than 2. This is an extraneous solution that must be discarded.

Final Answer: The only solution is $x=4$.

Question 3: Number of Solutions to a Logarithmic Equation

The Question: The number of solutions of $\ln(\frac{x^2}{4}) - \ln(x-1) = \log_5 5$ is __________.

• 0
• 1
• 2
• 3

Detailed Solution:

1. Determine the Domain:

   • For $\ln(\frac{x^2}{4})$, we need $\frac{x^2}{4} > 0 \implies x^2 > 0 \implies x \neq 0$.
   • For $\ln(x-1)$, we need $x-1 > 0 \implies x > 1$.
   • To satisfy both, the domain of the equation is $x > 1$.

2. Simplify the Equation:

   • The right side is $\log_5 5 = 1$.
   • The left side is a difference of two logs with the same base ($e$). We use the Quotient Rule.
   • $\ln\left(\frac{x^2/4}{x-1}\right) = 1$
   • $\ln\left(\frac{x^2}{4(x-1)}\right) = 1$

3. Convert to Exponential Form: The base of $\ln$ is $e$.

   • $e^1 = \frac{x^2}{4(x-1)}$

4. Solve for x:

   • $e = \frac{x^2}{4x - 4}$
   • $e(4x - 4) = x^2$
   • $4ex - 4e = x^2$
   • Rearrange into a standard quadratic equation: $x^2 - 4ex + 4e = 0$.

5. Check for Real Solutions using the Discriminant:

   • For this quadratic, $a=1, b=-4e, c=4e$.
   • The discriminant is $\Delta = b^2 - 4ac = (-4e)^2 - 4(1)(4e) = 16e^2 - 16e$.
   • We can factor this: $\Delta = 16e(e-1)$.
   • Since $e \approx 2.718$, both $16e$ and $(e-1)$ are positive numbers. Therefore, the discriminant $\Delta$ is positive.
   • A positive discriminant means there are two distinct real roots.

6. Check if the roots are in the domain ($x>1$):

   • The sum of the roots is $-b/a = 4e > 1$. The product of the roots is $c/a = 4e > 1$.
   • Since the sum and product are both positive, both roots must be positive.
   • We can also show that both roots are greater than 1 (as shown in the thought process). Both solutions are valid.

Final Answer: The number of solutions is 2.