If 1, log _9(3^{1-x}+2), log _3(4 cdot 3^x-1) | vedclass

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

Question

(B) Given that $1, \log _9(3^{1-x}+2), \log _3(4 \cdot 3^x-1)$ are in $A.P.$Since $2b = a + c$ for an $A.P.$, we have:$2 \log _9(3^{1-x}+2) = 1 + \log

Vedclass · Accepted Answer

$1-\log _3 4$

Vedclass · Answer

(B) Given that $1, \log _9(3^{1-x}+2), \log _3(4 \cdot 3^x-1)$ are in $A.P.$Since $2b = a + c$ for an $A.P.$, we have:$2 \log _9(3^{1-x}+2) = 1 + \log _3(4 \cdot 3^x-1)$Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$, we get $\log_9(y) = \frac{1}{2} \log_3(y)$:$2 \cdot \frac{1}{2} \log _3(3^{1-x}+2) = \log _3 3 + \log _3(4 \cdot 3^x-1)$$\log _3(3^{1-x}+2) = \log _3(3(4 \cdot 3^x-1))$$3^{1-x}+2 = 12 \cdot 3^x - 3$Let $3^x = t$. Then $\frac{3}{t} + 2 = 12t - 3$$3 + 2t = 12t^2 - 3t$$12t^2 - 5t - 3 = 0$$(4t - 3)(3t + 1) = 0$Since $t = 3^x > 0$, we have $t = \frac{3}{4}$.$3^x = \frac{3}{4} \Rightarrow x = \log_3 \left(\frac{3}{4}\right) = \log_3 3 - \log_3 4 = 1 - \log_3 4$.

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 sequence $\log a, \log \frac{a^2}{b}, \log \frac{a^3}{b^2}, \ldots$ is

If twice the $11^{th}$ term of an $A.P.$ is equal to $7$ times its $21^{st}$ term,then its $25^{th}$ term is equal to

If the sum of $n$ terms of an $A.P.$ is $3n^2 + 5n$ and $T_m = 164$,then $m = $

The four arithmetic means between $3$ and $23$ are.....

$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped the second day,$4$ more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is

Vedclass Test Series

Exam Paper Generator

Online Exam Module