The value of 6th term from the beggining of 2log 2-9x-1-7-1-21-5log 23x-1-17-84 then value of x isA- 0B- 1C- 2D- 3

The correct options are B 1 C 2T_5+1=^7C_5(2^log_2√(9^x 1+7))^2(12^1/5log_23^x 1+1)^5 84=7.3(2^log_29^x 1+7)(12^log_23^x 1+1) 4=2^(log_2(9^x 1+7) log_2(3^x 1+ ...

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Byju's Answer

Standard XII

Mathematics

Greatest Binomial Coefficients

The value of ...

Question

The value of $6^{th}$ term from the beggining of ${(2^{{log}_{2} \sqrt{9^{x - 1} + 7}} + \frac{1}{2^{\frac{1}{5} {log}_{2} (3^{x - 1} + 1)}})}^{7} = 84$ then value of $x$ is

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Solution

The correct options are
B $1$
C $2$
$T_{5 + 1} =^{7} C_{5} (2^{{log}_{2} \sqrt{9^{x - 1} + 7}})^{2} {(\frac{1}{2^{\frac{1}{5} {log}_{2} 3^{x - 1} + 1}})}^{5} 84 = 7.3 (2^{{log}_{2} 9^{x - 1} + 7}) (\frac{1}{2^{{log}_{2} 3^{x - 1} + 1}}) 4 = 2^{({log}_{2} (9^{x - 1} + 7) - {log}_{2} (3^{x - 1} + 1)} 4 = 2^{{log}_{2} \frac{(9^{x - 1} + 7)}{(3^{x - 1} + 1)}} 4 = \frac{(9^{x - 1} + 7)}{(3^{x - 1} + 1)} 4 (3^{x - 1} + 1) = (9^{x - 1} + 7) 3^{2 (x - 1)} - {4.3}^{x - 1} + 3 = 0 (3^{(x - 1)} - 1) (3^{(x - 1)} - 3) = 0 \Rightarrow x = 1, 2$

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Similar questions

Q. The value of

x

, for which the

6 t h

term in the expansion

{⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2^{{log}_{2} \sqrt{(9^{x - 1} + 7)}} + \frac{1}{2^{⎛ ⎝ \frac{1}{5} ⎞ ⎠} {log}_{2} (3^{x - 1} + 1)} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭}^{7}

84

, is equal to

Q. In the expansion of

{⎡ ⎣ 2^{{log}_{2} \sqrt{9^{x - 1} + 7}} + \frac{1}{2^{\frac{1}{5} {log}_{2} (3^{x - 1} + 1)}} ⎤ ⎦}^{7}

, 6th term is

84

. Then

x =

Q. The value of

x

, for which the

6 t h

term in the expansion

{2^{{log}_{2} \sqrt{(9^{x - 1} + 7)}} + \frac{1}{2^{\frac{1}{5} {log}_{2} (3^{x - 1} + 1)}}}^{7} is 84

, is equal to

Q. Determine the value of x for which the 6th term in

(2^{l o g_{2} (\sqrt{g^{x - y} + 7})} + \frac{1}{2^{\frac{1}{5} l o g_{2} (3^{x - 1} + 1)}})

is 84.

Q. The value of x for which the sixth term in the expansion of

{[2^{{log}_{2} \sqrt{9^{x - 1} + 7}} + \frac{1}{2^{\frac{1}{5} {log}_{2} (3^{x - 1} + 1)}}]}^{7}

is equal to 84 is

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The value of 6th term from the beggining of (2log2√9x−1+7+1215log2(3x−1+1))7=84 then value of x is

The correct options are B 1 C 2T5+1=7C5(2log2√9x−1+7)2(1215log23x−1+1)584=7.3(2log29x−1+7)(12log23x−1+1)4=2(log2(9x−1+7)−log2(3x−1+1)4=2log2(9x−1+7)(3x−1+1)4=(9x−1+7)(3x−1+1)4(3x−1+1)=(9x−1+7)32(x−1)−4.3x−1+3=0(3(x−1)−1)(3(x−1)−3)=0⇒x=1,2

The value of $6^{th}$ term from the beggining of ${(2^{{log}_{2} \sqrt{9^{x - 1} + 7}} + \frac{1}{2^{\frac{1}{5} {log}_{2} (3^{x - 1} + 1)}})}^{7} = 84$ then value of $x$ is