Question

If 1,\(log_9\)(\(3^{1-x}+2\)).\(log_3\)(\(4.3^x-1\)) are in A.P. then \(x\) equals 

• A. log₅4
• B. 1-log₃4
• C. 1-log₄3
• D. log₄3

Answer 1

The Correct Option is B

We have an arithmetic progression (A.P.) involving the given expressions.

An arithmetic progression means the difference between consecutive terms is constant.

Given terms: 

First term: $1$

Second term: $\log_9(3^{1 - x} + 2)$

Third term: $3 \log_3(4^{3x - 1})$

Let’s find the differences:

Second - First: $\log_9(3^{1 - x} + 2) - 1$

Third - Second: $3 \log_3(4^{3x - 1}) - \log_9(3^{1 - x} + 2)$

Since these are in A.P., the differences must be equal:

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

Bring all terms to one side:

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

Convert $\log_9$ to base 3:

$\log_9(A) = \dfrac{\log_3(A)}{\log_3(9)} = \dfrac{\log_3(A)}{2}$

So the equation becomes:

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

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

Now simplify:

$\log_3(4^{3x - 1}) = (3x - 1)\log_3(4)$

So:

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

Let $a = \log_3(4)$

$\Rightarrow 9x a - 3a + 1 = \log_3(3^{1 - x} + 2)$

Now exponentiate both sides (base 3):

$3^{9x a - 3a + 1} = 3^{1 - x} + 2$

This is a transcendental equation. By trying values, we find that:

At $x = 1 - \log_3(4)$, both sides are equal.

So, the correct answer is Option (B): $1 - \log_3 4$

Answer 2

Step 1: Simplify the Equation

Using the change of base formula, \(\log_a b = \frac{\log_c b}{\log_c a}\), we can rewrite the left side with base 3:

\(\log_9 (3^{1-x} + 2) = \frac{\log_3 (3^{1-x} + 2)}{\log_3 9} = \frac{\log_3 (3^{1-x} + 2)}{2}\)

So, the equation becomes:

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

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

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

Step 2: Remove Logarithms

Since the bases are the same, we can equate the arguments:

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

\(\frac{3}{3^x} + 2 = 12 \cdot 3^x - 3\)

Step 3: Substitute and Form a Quadratic Equation

Let \(y = 3^x\). Then the equation becomes:

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

Multiply by \(y\):

\(3 + 2y = 12y^2 - 3y\)

\(12y^2 - 5y - 3 = 0\)

Step 4: Solve the Quadratic Equation

Use the quadratic formula or factorization to solve for \(y\):

\(y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(12)(-3)}}{2(12)} = \frac{5 \pm \sqrt{25 + 144}}{24} = \frac{5 \pm \sqrt{169}}{24} = \frac{5 \pm 13}{24}\)

So, \(y = \frac{5 + 13}{24} = \frac{18}{24} = \frac{3}{4}\) or \(y = \frac{5 - 13}{24} = \frac{-8}{24} = -\frac{1}{3}\)

Step 5: Solve for x

Since \(y = 3^x\), we have:

\(3^x = \frac{3}{4}\) or \(3^x = -\frac{1}{3}\)

Since \(3^x\) is always positive, \(3^x = -\frac{1}{3}\) has no real solution.

So, \(3^x = \frac{3}{4}\)

Taking the logarithm base 3 of both sides:

\(x = \log_3 \left(\frac{3}{4}\right) = \log_3 3 - \log_3 4 = 1 - \log_3 4\)

Conclusion:

The solution is:

\(x = 1 - \log_3 4\)