JAMIA 2022 — Mathematics PYQ

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Question

JAMIA 2022 — Mathematics PYQ

JAMIA | Mathematics | 2022

If 1, log₉ (3^1-x+ 2) and log₃(4.3^x –1) are in A.P., then x equals

Choose the correct answer:

Explanation

1. Set Up the Equation

The given terms are $1$ , $\log_9(3^{1-x} + 2)$ , and $\log_3(4 \cdot 3^x - 1)$ .

Using the A.P. property:

2 \log_9(3^{1-x} + 2) = 1 + \log_3(4 \cdot 3^x - 1)

2. Simplify the Logarithms

First, we change the base of the left side. Since $9 = 3^2$ , we use the property $\log_{a^n}(b) = \frac{1}{n}\log_a(b)$ :

2 \cdot \frac{1}{2} \log_3(3^{1-x} + 2) = 1 + \log_3(4 \cdot 3^x - 1)

\log_3(3^{1-x} + 2) = \log_3(3) + \log_3(4 \cdot 3^x - 1)

Using the logarithmic addition property $\log(m) + \log(n) = \log(mn)$ :

\log_3(3^{1-x} + 2) = \log_3(3(4 \cdot 3^x - 1))

Now, remove the logarithms from both sides:

3^{1-x} + 2 = 3(4 \cdot 3^x - 1)

3. Solve the Exponential Equation

Let $3^x = y$ . Then $3^{1-x} = \frac{3}{3^x} = \frac{3}{y}$ .

Substitute $y$ into the equation:

\frac{3}{y} + 2 = 3(4y - 1)

\frac{3}{y} + 2 = 12y - 3

Multiply the entire equation by $y$ to clear the fraction:

3 + 2y = 12y^2 - 3y

12y^2 - 5y - 3 = 0

4. Solve the Quadratic Equation

Using the quadratic formula or factoring:

12y^2 - 9y + 4y - 3 = 0

3y(4y - 3) + 1(4y - 3) = 0

(3y + 1)(4y - 3) = 0

Possible values for $y$ :

$y = -\frac{1}{3}$ (Not possible since $3^x$ must be positive)
$y = \frac{3}{4}$

5. Find $x$

Since $3^x = y$ :

3^x = \frac{3}{4}

Taking $\log_3$ on both sides:

x = \log_3\left(\frac{3}{4}\right)

x = \log_3(3) - \log_3(4)

x = 1 - \log_3(4)