The sum of the roots of equation log 232-x-6x-3-x log 23-2A- 3B- 7C- 9D- 5

The correct option is A 3log_2(3^2+x 6^x) = 3 + xlog_2(32) ⇒log_2(3^2+x 6^x(32)^x) = 3 ⇒2^x· 3^x(9 2^x)3^x=8 ⇒ 2^x(9 2^x) = 8 Assuming 2^x = t, we get ⇒ ...

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Standard XIII

Mathematics

Roots of a Quadratic Equation

The sum of th...

Question

The sum of the roots of equation ${log}_{2} (3^{2 + x} - 6^{x}) = 3 + x {log}_{2} (\frac{3}{2})$ is

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Solution

The correct option is A $3$
${log}_{2} (3^{2 + x} - 6^{x}) = 3 + x {log}_{2} (\frac{3}{2})$
$\Rightarrow {log}_{2} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{3^{2 + x} - 6^{x}}{{(\frac{3}{2})}^{x}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = 3 \Rightarrow \frac{2^{x} \cdot 3^{x} (9 - 2^{x})}{3^{x}} = 8 \Rightarrow 2^{x} (9 - 2^{x}) = 8$
Assuming $2^{x} = t$ , we get
$\Rightarrow t (9 - t) = 8 \Rightarrow t^{2} - 9 t + 8 = 0 \Rightarrow (t - 8) (t - 1) = 0 \Rightarrow t = 1, 8 \Rightarrow 2^{x} = 1, 8 \Rightarrow x = 0, 3 ∴ Sum = 0 + 3 = 3$

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The sum of the roots of equation log2(32+x−6x)=3+xlog2(32) is

The correct option is A 3log2(32+x−6x)=3+xlog2(32) ⇒log2⎛⎜ ⎜ ⎜ ⎜⎝32+x−6x(32)x⎞⎟ ⎟ ⎟ ⎟⎠=3⇒2x⋅3x(9−2x)3x=8⇒2x(9−2x)=8 Assuming 2x=t, we get ⇒t(9−t)=8⇒t2−9t+8=0⇒(t−8)(t−1)=0⇒t=1,8⇒2x=1,8⇒x=0,3∴Sum=0+3=3

The sum of the roots of equation ${log}_{2} (3^{2 + x} - 6^{x}) = 3 + x {log}_{2} (\frac{3}{2})$ is