Question

Please answer my question .
Don't Spam it is for class 11
All steps are needed​

Answer 1

Step-by-step explanation:

We have,

$1\: , \:\log_{9}( {3}^{1 - x} + 2 ) \: ,\: \log_{3}(4. {3}^{x} - 1 )$ are in A.P.

So,

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

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

$\implies \: 2 \log_{ {3}^{2} }( {3}^{1 - x} + 2 ) = \log_{3} \{3(4. {3}^{x} - 1 ) \}$

$\implies \: 2 \times \frac{1}{2} \log_{ 3 }( {3}^{1 - x} + 2 ) = \log_{3} \{4. {3}^{x + 1} - 3\}$

$\implies \:\log_{ 3 }( {3}^{1 - x} + 2 ) = \log_{3} \{4. {3}^{x + 1} - 3\}$

$\implies \:( {3}^{1 - x} + 2 ) = (4. {3}^{x + 1} - 3)$

$\implies \: \frac{3}{ {3}^{x} } + 2 = 12( {3}^{x}) - 3$

$\implies \: \frac{3}{ {3}^{x} } = 12( {3}^{x}) -5$

$\implies \: 3= 12( {3}^{x})({3}^{x})-5( {3}^{x} )$

$\implies \: 12( {3}^{x})^{2} -5( {3}^{x} ) - 3 = 0$

$\implies \: 12( {3}^{x})^{2} -9( {3}^{x} ) + 4( {3}^{x}) - 3 = 0$

$\implies \: 3( {3}^{x}) \{4 ( {3}^{x}) -3 \} + 1 \{4( {3}^{x}) - 3 \}= 0$

$\implies \: \{3( {3}^{x}) + 1 \} \{4 ( {3}^{x}) -3 \}= 0$

$\implies \: 3( {3}^{x}) + 1 = 0 \: \: or \: \: 4 ( {3}^{x}) -3 = 0$

$\implies \: {3}^{x} = - \frac{1}{3} \: \: or \: \: {3}^{x} = \frac{3}{4}$

Since, exponential function is always positive, so, it can't take negative values..

So,

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

$\implies \log_{3}( {3}^{x}) = \log_{3} \bigg( \frac{3}{4} \bigg)$

$\implies x\log_{3}( 3) = \log_{3} \bigg( \frac{3}{4} \bigg)$

$\implies x = 1 - \log_{3} ( 4)$

$\sf{\huge{\mathbb{PROPERTY\:\:USED:}}}$

$\large\dagger \: \log_{ {a}^{n} }(b) = \frac{1}{n} \log_{a}(b)$

$\large\dagger \: \log_{ a }(b) ^{n} = n \log_{a}(b)$

$\large\dagger \: \log_{ a }(b) + \log_{a}(c) = \log_{a}(bc)$

$\large\dagger \: \log_{ a }(b) - \log_{a}(c) = \log_{a} \bigg( \frac{b}{c} \bigg)$