GO Classes CS/DA 2025 | Weekly Quiz 3 | Fundamental Course and Linear Algebra | Question: 13

Answer 1

$\begin{aligned}
& \sqrt{4 \log _2 x}+4 \log _4\left(\frac{2}{x}\right)^{1 / 2}=2 \\
& 2 \sqrt{\log _2 x}+\frac{4}{2} \log \left(\frac{2}{x}\right)=2 \\
& 2 \sqrt{\log _2 x}+2 \log \left(\frac{2}{x}\right)=2 \\
& \sqrt{\log _2 x}+\log _4 2-\log _4 x=1 \\\end{aligned}$

$\begin{aligned}& \sqrt{\log _2 x}+\frac{\log 2}{\log 4}-\frac{\log x}{\log 4}=1 \\
& \sqrt{\log _2 x}+\frac{\log 2}{\log 2^2}-\frac{\log x}{\log 2^2}=1 \\
& \sqrt{\log _2 x}+\frac{\log 2}{2 \log 2}-\frac{\log x}{2 \log 2}=1 \\ \end{aligned}$

$\begin{aligned}& \sqrt{\log x}+\frac{1}{2}-\frac{1}{2} \log x=1 \\ & \sqrt{\log _2 x}-\frac{1}{2} \log _2 x=\frac{1}{2} \end{aligned}$

$\text{Assume }\log_2 x = m$

$\sqrt{m} - \frac{1}{2}m = \frac{1}{2}$

$\begin{aligned}& \text { SQUARING } \\ & 4 m=1+2 m+m^2 \\ & m^2-2 m+1=0 \\ & (m-1)^2=0 \\ \end{aligned}$

$\begin{aligned}&m-1=0 \\ & m=1 \\ & \log _2 x=1 \\ & \boxed{x=2} \\ & \end{aligned}$

Comments

just a minor formatting issue here
 

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i correctly followed till step m=1 but forogt to put value of m return to Log2X. and submitted the answer as 1.  :(

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Easiest way to solve it !!!

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@pranav25 Your step 3 is incorrect. Since you cannot directly remove the square root on the log by taking the square root of the function inside the log.

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Thank you for correcting me !!! can you please to tell me the correct way of doing so?

Answer 2

$\sqrt{\log_2{x^4}} + 4\log_4{\sqrt{2/x}} = 2$

$\Rightarrow$ $\sqrt{4\log_2{x}} + \frac{4}4{\log_2{(\frac{2}x)}} = 2$

$\Rightarrow$ $2\sqrt{\log_2{x}} + \log_2{2} - \log_2{x} = 2$

$\Rightarrow$ $2\sqrt{\log_2{x}} - \log_2{x} = 1$

$\Rightarrow$ $2\sqrt{\log_2{x}} = 1 + \log_2{x}$

Squaring both sides,

$\Rightarrow$ $4\log_2{x} = 1 + (\log_2{x})^2 + 2\log_2{x}$

$\Rightarrow$ $(\log_2{x})^2 - 2\log_2{x} + 1 = 0$

$\Rightarrow$ $(\log_2{x}-1)^2 = 0$

$\Rightarrow$ $\log_2{x}= 1$

$\Rightarrow$ $x=2$

Comments

Hello @sankalps, in line 2 of your derivation, it will be 4/2 instead of 4/4 in the second term. 

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It will be 4/4 only. One 2 in the denominator is from the power of 2/x and the other 2 is from the base (base is 4 and in the next line I changed it to 2)

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Oh yes yes, I got it now. thanks.

Answer 3

a

Answer 4

For log₂ (x⁴)  and log₄ (2/x)^0.5   :-    x⁴ > 0  and  (2/x)^0.5 > 0   =>  x > 0