If 1, log_9(3^{1-x} + 2), log_3(4 cdot 3^x - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Since the given numbers are in $A.P.$, the middle term is the arithmetic mean of the other two terms.$2 \log_9(3^{1-x} + 2) = \log_3(4 \cdot 3^x -

Vedclass · Accepted Answer

$1 - \log_3 4$

Vedclass · Answer

(B) Since the given numbers are in $A.P.$, the middle term is the arithmetic mean of the other two terms.$2 \log_9(3^{1-x} + 2) = \log_3(4 \cdot 3^x - 1) + 1$Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$, we get:$2 \cdot \frac{1}{2} \log_3(3^{1-x} + 2) = \log_3(4 \cdot 3^x - 1) + \log_3 3$$\log_3(3^{1-x} + 2) = \log_3(3(4 \cdot 3^x - 1))$$3^{1-x} + 2 = 12 \cdot 3^x - 3$Let $y = 3^x$. Then $3^{1-x} = \frac{3}{y}$.$\frac{3}{y} + 2 = 12y - 3$$3 + 2y = 12y^2 - 3y$$12y^2 - 5y - 3 = 0$$(4y - 3)(3y + 1) = 0$Since $y = 3^x > 0$, we have $y = \frac{3}{4}$.$3^x = \frac{3}{4} \implies x = \log_3 \left(\frac{3}{4}\right) = \log_3 3 - \log_3 4 = 1 - \log_3 4$.

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

Similar Questions

The $G.M.$ of the roots of the equation $x^2 - 18x + 9 = 0$ is

The $p^{th}$ term of the series $\left( 3 - \frac{1}{n} \right) + \left( 3 - \frac{2}{n} \right) + \left( 3 - \frac{3}{n} \right) + \dots$ will be

The sum of the series $(1^2 + 1) \cdot 1! + (2^2 + 1) \cdot 2! + (3^2 + 1) \cdot 3! + \dots + (n^2 + 1) \cdot n!$ is:

If the sum of the first $15$ terms of the series ${\left( {\frac{3}{4}} \right)^3} + {\left( {1\frac{1}{2}} \right)^3} + {\left( {2\frac{1}{4}} \right)^3} + {3^3} + {\left( {3\frac{3}{4}} \right)^3} + \dots$ is equal to $225k$,then $k$ is equal to

If the sum of the first $40$ terms of the series $3+4+8+9+13+14+18+19+\ldots$ is $(102)m$,then $m$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module