If $1, \log_9 (3^{1-x} + 2), \log_3 (4 \cdot 3^x - 1)$ are in A.P. then $x$ equals Options: (A) $\log_3 4$ (B) $1 - \log_3 4$ (C) $1 - \log_4 3$ (D) $\log_4 3$

### Analyzing the Setup Imagine you are standing at the threshold of a logarithmic landscape. You have three terms: $1$, $\log_9 (3^{1-x} + 2)$, and $\log_3 (4 \cdot 3^x - 1)$. These terms are locked in an Arithmetic Progression (A.P.). In any A.P., the middle term is the bridge between the first and the last. Mathematically, if $a, b, c$ are in A.P., then $2b = a + c$. Applying this to our terms, we obtain the following relationship: $$2 \log_9 (3^{1-x} + 2) = 1 + \log_3 (4 \cdot 3^x - 1)$$ ### The Logarithmic Jungle We observe a base mismatch between $9$ and $3$. We utilize the change-of-base property, $\log_{a^k} B = \frac{1}{k} \log_a B$. Since $9 = 3^2$, we rewrite the term as $\log_9 (3^{1-x} + 2) = \frac{1}{2} \log_3 (3^{1-x} + 2)$. When substituted back into our equation, the $2$ and the $\frac{1}{2}$ cancel each other out, simplifying the expression to: $$\log_3 (3^{1-x} + 2) = 1 + \log_3 (4 \cdot 3^x - 1)$$ ### The Quadratic Transformation To combine the terms, we express the constant $1$ as $\log_3 3$. The equation now reads: $$\log_3 (3^{1-x} + 2) = \log_3 3 + \log_3 (4 \cdot 3^x - 1)$$ Using the product rule $\log A + \log B = \log(AB)$, the right side becomes $\log_3 [3 \cdot (4 \cdot 3^x - 1)]$, which simplifies to $\log_3 (12 \cdot 3^x - 3)$. Equating the arguments of the logarithms, we get: $$3^{1-x} + 2 = 12 \cdot 3^x - 3$$ Let $t = 3^x$. The equation transforms into: $$\frac{3}{t} + 2 = 12t - 3$$ Multiplying by $t$ yields the quadratic equation: $$12t^2 - 5t - 3 = 0$$ ### The Final Victory Solving $12t^2 - 5t - 3 = 0$ by splitting the middle term gives us the factors $(4t - 3)(3t + 1) = 0$. This results in two potential values: $t = \frac{3}{4}$ or $t = -\frac{1}{3}$. We must respect the domain constraints. Since $t = 3^x$, $t$ must be strictly positive. Therefore, we reject $t = -\frac{1}{3}$. Setting $3^x = \frac{3}{4}$ and taking the logarithm on both sides, we find: $$x = \log_3 \left( \frac{3}{4} \right)$$ This simplifies to $x = \log_3 3 - \log_3 4$. Thus, the final solution is: **$x = 1 - \log_3 4$**

JEE Main 2002

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Animated Solution for Mathematics - Sequence and Series: If $1, lo g_{9} (3^{1 - x} + 2), lo g_{3} (4 \cdot 3^{x} - 1)$ are in A.P. then $x$ equals

Q: Why do we reject the negative root for $t$?

Great catch! We substituted $t = 3^x$. Since $3$ is a positive base, $3^x$ is strictly positive for all real $x$. A negative value for $t$ would imply $3^x = -\frac{1}{3}$, which is impossible in the real number system. Always check the domain of your substitution!

Q: Is there a faster way to handle the $\log_9$ term?

You could convert everything to base $9$, but that would make the right side messy. Converting to base $3$ is almost always the cleaner path because $9$ is a power of $3$. Always look for the 'base' that simplifies the most terms.

Select Answer:

Progressive Hints

Hint 1

The sequence is in Arithmetic Progression. Before you dive into the logs, remember the fundamental property: for any three terms

a, b, c

in A.P., the middle term is the arithmetic mean of the others, expressed as

2 b = a + c

. The trap here isn't the A.P. itself, but the hidden base mismatch in your logarithmic terms.

Hint 2

You have a base

9

and a base

3

. You cannot combine these directly. Use the change-of-base property:

lo g_{a^{k}} B = \frac{1}{k} lo g_{a} B

. Once you normalize the bases to

3

, you can use the product rule

lo g A + lo g B = lo g (A B)

to simplify the right-hand side.

Hint 3

After simplifying, you will have an equation involving

3^{x}

. To break the mental block, substitute

t = 3^{x}

. This will transform your logarithmic equation into a standard quadratic equation in

t

. Solve for

t

, but remember: an exponential function

3^{x}

can never be negative. Reject any invalid roots!

Visualized Solution

Arithmetic Progression Condition

Given sequence: $1, lo g_{9} (3^{1 - x} + 2), lo g_{3} (4 \cdot 3^{x} - 1)$
These terms are in Arithmetic Progression (A.P.).
For three terms $a, b, c$ in A.P., the property is: $2 b = a + c$

Applying the A.P. Formula

Here, $a = 1$ , $b = lo g_{9} (3^{1 - x} + 2)$ , and $c = lo g_{3} (4 \cdot 3^{x} - 1)$
Substituting into $2 b = a + c$ :
$2 lo g_{9} (3^{1 - x} + 2) = 1 + lo g_{3} (4 \cdot 3^{x} - 1)$

Simplifying the Base

Notice the bases are $9$ and $3$ . We need a common base.
Using the property: $lo g_{a^{k}} B = \frac{1}{k} lo g_{a} B$
Since $9 = 3^{2}$ , we have: $lo g_{9} (3^{1 - x} + 2) = \frac{1}{2} lo g_{3} (3^{1 - x} + 2)$

Canceling the Coefficient

Substituting back into our equation:
$2 \cdot \frac{1}{2} lo g_{3} (3^{1 - x} + 2) = 1 + lo g_{3} (4 \cdot 3^{x} - 1)$
The $2$ and $\frac{1}{2}$ cancel out.
Simplified equation: $lo g_{3} (3^{1 - x} + 2) = 1 + lo g_{3} (4 \cdot 3^{x} - 1)$

Expressing Constant as Logarithm

To combine terms, express $1$ as a logarithm with base $3$ .
$1 = lo g_{3} 3$
The equation becomes: $lo g_{3} (3^{1 - x} + 2) = lo g_{3} 3 + lo g_{3} (4 \cdot 3^{x} - 1)$

Product Rule of Logarithms

Using the property $lo g A + lo g B = lo g (A B)$ :
$lo g_{3} (3^{1 - x} + 2) = lo g_{3} [3 \cdot (4 \cdot 3^{x} - 1)]$
$lo g_{3} (3^{1 - x} + 2) = lo g_{3} (12 \cdot 3^{x} - 3)$

Equating Log Arguments

Since the bases are the same, we can equate the arguments:
$3^{1 - x} + 2 = 12 \cdot 3^{x} - 3$
Rewrite $3^{1 - x}$ using exponent rules: $3^{1 - x} = \frac{3}{3 ^{x}}$
$\frac{3}{3 ^{x}} + 2 = 12 \cdot 3^{x} - 3$

Substitution: Let $3^{x} = t$

To make it easier to solve, let $3^{x} = t$ .
The equation becomes: $\frac{3}{t} + 2 = 12 t - 3$
Rearranging the constant terms: $\frac{3}{t} + 5 = 12 t$

Forming the Quadratic

Multiply the entire equation by $t$ to remove the fraction:
$3 + 5 t = 12 t^{2}$
Rearranging into standard quadratic form:
$12 t^{2} - 5 t - 3 = 0$

Factoring the Quadratic

We need two numbers that multiply to $- 36$ and add to $- 5$ . These are $- 9$ and $4$ .
Split the middle term: $12 t^{2} - 9 t + 4 t - 3 = 0$
Factor by grouping: $3 t (4 t - 3) + 1 (4 t - 3) = 0$
$(4 t - 3) (3 t + 1) = 0$
Possible values: $t = \frac{3}{4}$ or $t = - \frac{1}{3}$

Domain Constraint

Recall our substitution: $t = 3^{x}$
An exponential function $a^{x}$ is always strictly positive ( $> 0$ ).
Therefore, $3^{x} = - \frac{1}{3}$ is rejected.
We must take $t = \frac{3}{4}$ .

Finding the Value of $x$

We have: $3^{x} = \frac{3}{4}$
Taking $lo g_{3}$ on both sides: $x = lo g_{3} (\frac{3}{4})$
Using the quotient rule $lo g \frac{A}{B} = lo g A - lo g B$ :
$x = lo g_{3} 3 - lo g_{3} 4$
$x = 1 - lo g_{3} 4$

Final Answer

The value of $x$ is $1 - lo g_{3} 4$ .
This matches the second option provided in the question.

00:00 / 00:00

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

Imagine you are standing at the threshold of a logarithmic landscape. You have three terms:

1

lo g_{9} (3^{1 - x} + 2)

, and

lo g_{3} (4 \cdot 3^{x} - 1)

These terms are locked in an Arithmetic Progression (A.P.). In any A.P., the middle term is the bridge between the first and the last.

Mathematically, if

a, b, c

are in A.P., then

2 b = a + c

. Applying this to our terms, we obtain the following relationship:

2 lo g_{9} (3^{1 - x} + 2) = 1 + lo g_{3} (4 \cdot 3^{x} - 1)

The Logarithmic Jungle

We observe a base mismatch between

9

and

3

. We utilize the change-of-base property,

lo g_{a^{k}} B = \frac{1}{k} lo g_{a} B

Since

9 = 3^{2}

, we rewrite the term as

lo g_{9} (3^{1 - x} + 2) = \frac{1}{2} lo g_{3} (3^{1 - x} + 2)

When substituted back into our equation, the

2

and the

\frac{1}{2}

cancel each other out, simplifying the expression to:

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

The Quadratic Transformation

To combine the terms, we express the constant

1

lo g_{3} 3

. The equation now reads:

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

Using the product rule

lo g A + lo g B = lo g (A B)

, the right side becomes

lo g_{3} [3 \cdot (4 \cdot 3^{x} - 1)]

, which simplifies to

lo g_{3} (12 \cdot 3^{x} - 3)

Equating the arguments of the logarithms, we get:

3^{1 - x} + 2 = 12 \cdot 3^{x} - 3

Let

t = 3^{x}

. The equation transforms into:

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

Multiplying by

t

yields the quadratic equation:

12 t^{2} - 5 t - 3 = 0

The Final Victory

Solving

12 t^{2} - 5 t - 3 = 0

by splitting the middle term gives us the factors

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

. This results in two potential values:

t = \frac{3}{4}

t = - \frac{1}{3}

We must respect the domain constraints. Since

t = 3^{x}

t

must be strictly positive. Therefore, we reject

t = - \frac{1}{3}

Setting

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

and taking the logarithm on both sides, we find:

x = lo g_{3} (\frac{3}{4})

This simplifies to

x = lo g_{3} 3 - lo g_{3} 4

. Thus, the final solution is:

$x = 1 - lo g_{3} 4$

Common Traps & FAQs

Wait, why can't we just equate the terms directly? Why do we need

2 b = a + c

Equating terms directly would imply the sequence is constant, which isn't necessarily true! An A.P. just requires a constant difference. The condition

2 b = a + c

is the algebraic definition of that constant difference, ensuring the gap between the first and second term equals the gap between the second and third.

Why do we reject the negative root for

t

Great catch! We substituted

t = 3^{x}

. Since

3

is a positive base,

3^{x}

is strictly positive for all real

x

. A negative value for

t

would imply

3^{x} = - \frac{1}{3}

, which is impossible in the real number system. Always check the domain of your substitution!

Is there a faster way to handle the

lo g_{9}

term?

You could convert everything to base

9

, but that would make the right side messy. Converting to base

3

is almost always the cleaner path because

9

is a power of

3

. Always look for the 'base' that simplifies the most terms.

Animated Solution for Mathematics - Sequence and Series: If $1, lo g_{9} (3^{1 - x} + 2), lo g_{3} (4 \cdot 3^{x} - 1)$ are in A.P. then $x$ equals

Select Answer:

Visualized Solution

Arithmetic Progression Condition

Applying the A.P. Formula

Simplifying the Base

Canceling the Coefficient

Expressing Constant as Logarithm

Product Rule of Logarithms

Equating Log Arguments

Substitution: Let $3^{x} = t$

Forming the Quadratic

Factoring the Quadratic

Domain Constraint

Finding the Value of $x$

Final Answer

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Logarithmic Jungle

The Quadratic Transformation

The Final Victory

Similar Questions

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1,log9​(31−x+2),log3​(4⋅3x−1) are in A.P. then x equals

Select Answer:

Visualized Solution

Arithmetic Progression Condition

Applying the A.P. Formula

Simplifying the Base

Canceling the Coefficient

Expressing Constant as Logarithm

Product Rule of Logarithms

Equating Log Arguments

Substitution: Let .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3x=t

Forming the Quadratic

Factoring the Quadratic

Domain Constraint

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Final Answer

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Logarithmic Jungle

The Quadratic Transformation

The Final Victory

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