Question

Plz answer my question with all steps
Don't spam
All steps are needed​

Answer 1

$\\ \underline\bold{ \large \: Solution : -}$

We know that , If a,b,c are in A.P .The difference between two consecutive numbers are equal. That means 2b = a + c

So, Let

• $\: \sf \: a = 1$

• $\: \: \sf b = log_9( {3}^{1 - x} + 2)$

• $\sf \: \: c = log_{3}(4 \times {3}^{x} - 1)$

Therefore,

$\\ \sf \: \: 2 \times log_{9}( {3}^{1 - x} + 2) = log_{3}(4 \times {3}^{x} - 1) + 1$

$\\ \sf \qquad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dots using \: \bold{ log_{a}a = 1}$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: \: \: \: 2 \times log_{ {3}^{2} }( {3}^{1 - x} + 2) = log_{3}(4 \times {3}^{x} - 1) + log_{3}3$

$\\ \sf \qquad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dots using \: \log_{b}a + \log_{b}c = \log_bac$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: \: \: \: 2 \times \frac{1}{2} log_{ {3}}( {3}^{1 - x} + 2) = log_{3}(4 \times {3}^{x} - 1) 3$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: log_{ {3}}( {3}^{1 - x} + 2) = log_{3}(12\times {3}^{x} - 3)$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: {3}^{1 - x} + 2 = 12 \times {3}^{x} - 3$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: \frac{3}{ {3}^{x} } + 2 = 12 \times {3}^{x} - 3$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: 3 + 2 \times {3}^{x} = 12 {( {3}^{x} )}^{2} - 3 \times {3}^{x}$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: 12 \times {( {3 }^{x}) }^{2} - 5 \times {3}^{x} - 3 = 0$

Now , let's assume $\sf 3^x = t$ , so

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: 12{t }^{2} - 5 t - 3 = 0$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: 12{t }^{2} - 9t + 4t- 3 = 0$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: 3t(4t - 3) + 1(4t - 3)$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: (4t - 3) = 0 \qquad \: or \: \qquad (3t + 1) = 0$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: t = \frac{3}{4} \qquad \: or \: \qquad t = \bigg( - \frac{1}{3} \bigg)$

Now,

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: {3}^{x} = \frac{3}{4} \qquad \: or \: \qquad {3}^{x} = \bigg( - \frac{1}{3} \bigg)$

By applying log ,

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: \log {3}^{x} = \log \ \frac{3}{4} \qquad \: or \: \log {3}^{x} = \log\bigg( - \frac{1}{3} \bigg)$

$\\ \: \qquad \: \: \: \: \: \: \longmapsto\sf \: x\log {3}^ = \log \ \frac{3}{4} \qquad \: or \qquad\: x \log {3}^ = \log\bigg( - \frac{1}{3} \bigg)$

We know that , $\sf \log_ba$ can exists only a ≥ 0.

Therefore,

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: x \log 3 = \log \frac{3}{4}$

$\\ \sf \: \: \: \implies \: x \log3 = \log3 - \log4$

$\\ \sf \: \: \: \implies \: x = \frac{\log3 - \log4}{ \log3}$

$\\ \sf \: \: \: \implies \boxed{ \: \bold{ \: x = 1 - \log_{3}(4)}}$