Question

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Answer 1

Question :

If 1,  log_9 [ 3^( 1 - x ) + 2 ] and log_3 [ 4.3^x - 1 ] are in AP then the value of x is equal to

(A)  log_4 3

(B) log_3 4

(C) 1 - log_3 4

(D)  log_3 0.25

Answer :

Given :

1,  log_9 [ 3^( 1 - x ) + 2 ] and log_3 [ 4.3^x - 1 ] are in AP

We know that :

If a,  b,  c are in AP the relationship between them :

⇒ 2b = a + c

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

It can be written as

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

Since log_( a^m ) n = 1/m × log_a n

$\sf \implies 2 \bigg( \dfrac{1}{2} log_{ 3 }( {3}^{1 - x} + 2) \bigg) = log_{3}3+ log_{3}(4 \times 3^{x} - 1)$

Using Product rule log a + log b we get,

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

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

$\sf \implies \dfrac{3}{ {3}^{x} } + 2= 12(3^{x}) - 3$

Let us say 3^x = k and by substituting in the above equation we get,

$\sf \implies \dfrac{3}{ k} + 2= 12k - 3$

Multiplying every term with 'k' we get,

⇒ 3 + 2k = 12k² - 3k

⇒ 12k² - 5k - 3 = 0

Splitting the middle term

⇒ 12k² - 9k + 4k - 3 = 0

⇒ 3k( 4k - 3 ) + 1( 4k - 3 ) = 0

⇒ ( 3k + 1 )( 4k - 3 ) = 0

⇒ 3k + 1 = 0 (OR)  4k - 3 = 0

⇒ k = - 1/3 (OR)  k = 3/4

But k ≠ - 1/3 as 3^x can't be - 1/3

⇒ k = 3/4

⇒ 3^x = 3/4

⇒ x = log_3 ( 3/4 )

Since log ( a/b ) = log a - log b

⇒ x = log_3 3 - log_3 4

⇒ x = 1 - log_3 4

Therefore the value of x is equal to (C)  1 - log_3 4.