If the 6th term of 2log10(10−3x)+2(x−2)log1035m is 21 and if the coefficients of 2nd , 3rd and 4th are in A.P then x=

2 ( m C 2 ) = m C 1 + m C 3 1 ( m − 2 ) ! = 1 ( m − 1 ) ! + 1 6 ( m − 3 ) ! 1 m − 2 = 1 ( m − 1 ) ! + 1 6 1 m − 2 = 1 ( m − 1 ) ( m − 2 ) + 1 6 1 m − 2 1 − 1 m − 1 = 1 6 ( m − 2 ) ( m − 2 ) ( m − 1 ) = 1 6 ⇒ 6 ( m − 2 ) = ( m − 2 ) ( m − 1 ) ⇒ 6 ( m − 2 ) − ( m − 2 ) ( m − 1 ) = 6 ( m − 2 ) [ 6 − m + 1 ] = 6 m − 7 = 0 [ ∵ m ≠ 2 ] m = 7 Substitute m value in (1) we get 21 = 7 C 5 2 log 10 10 − 3 x 7 − 5 2 + log 10 3 x − 2 ⇒ log 10 10 − 3 x + log 10 3 x − 2 = 0 ⇒ log 10 ( 10 − 3 x ) ( 3 x − 2 ) = log 10 1 ⇒ ( 10 − 3 x ) 3 x 3 2 = 1 let 3 x = t ⇒ ( 10 − t ) ( t ) 9 = 1 ⇒ ( 10 − t ) t = 9 ⇒ t = 9 ( or ) 1 3 x = 1 = 3 0 x = 0 3 x = 9 = 3 2 x = 2 ∴ x = 0 or x = 2

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If the 6^th term of ${[\sqrt{2^{\log_{10} (10 - 3^{x})}} + \sqrt[5]{2^{(x - 2) \log_{10} 3}}]}^{m}$ is 21 and if the coefficients of 2^nd , 3^rd and 4^thare in A.P then $x =$

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Detailed Solution

$\begin{array}{l} 2 (^{m} C_{2}) =^{m} C_{1} +^{m} C_{3} \\ \frac{1}{(m - 2)!} = \frac{1}{(m - 1)!} + \frac{1}{6 (m - 3)!} \\ \frac{1}{m - 2} = \frac{1}{(m - 1)!} + \frac{1}{6} \\ \frac{1}{m - 2} = \frac{1}{(m - 1) (m - 2)} + \frac{1}{6} \\ \frac{1}{m - 2} [1 - \frac{1}{m - 1}] = \frac{1}{6} \\ \frac{(m - 2)}{(m - 2) (m - 1)} = \frac{1}{6} \\ \Rightarrow 6 (m - 2) = (m - 2) (m - 1) \\ \Rightarrow 6 (m - 2) - (m - 2) (m - 1) = 6 \\ (m - 2) [6 - m + 1] = 6 \\ m - 7 = 0 [∵ m \neq 2] \\ m = 7 \end{array}$

Substitute m value in (1) we get

$21 =^{7} C_{5} 2^{\log_{10} 10 - 3^{x} (\frac{7 - 5}{2}) + \log_{10} 3^{x - 2}}$

$\begin{array}{l} \Rightarrow \log_{10} 10 - 3^{x} + \log_{10} 3^{x - 2} = 0 \\ \Rightarrow \log_{10} (10 - 3^{x}) (3^{x - 2}) = \log_{10} 1 \\ \Rightarrow (10 - 3^{x}) (\frac{3^{x}}{3^{2}}) = 1 \\ let 3^{x} = t \\ \Rightarrow (10 - t) \frac{(t)}{9} = 1 \\ \Rightarrow (10 - t) t = 9 \\ \Rightarrow t = 9 (or) 1 \\ 3^{x} = 1 = 3^{0} \\ x = 0 \\ 3^{x} = 9 = 3^{2} \\ x = 2 \\ ∴ x = 0 or x = 2 \end{array}$