Finding the Freundlich adsorption isotherm from the graph of log(x/m) v/s log(p) The following problem was asked in JEE Mains 2020 (Sept 2, Shift 1),

The mass of gas adsorbed, $x$, per unit mass of adsorbate, $m$, was measured at various pressures, $p$. A graph between $\log\frac xm$ and $\log p$ gives a straight line with slope equal to $2$ and the intercept equal to $0.4771$. The value of $\frac xm$ at a pressure of $\pu{4 atm}$ is: (Given $\log 3 = 0.4771$)

I know the Freundlich adsorption isotherm formula, but I decided to write the equation of given line $( y = c + mx)$ as,
$$ \log \frac xm = \log 3 + 2\log p $$
So, the relation comes out to be,
$$ \frac xm = 3p^2 $$
On plugging $p = 4$, we get,
$$ \frac xm = 3(16) = 48 $$
But, the answer was wrong as per the official key, which mostly has genuine answers. Where am I wrong?

The answer given is,

 6
One version of Freundlich adsorption isotherm equation is:
$$ \frac xm = Kp^{\frac12}, $$
which can also be written as:
$$ \log \frac xm = \log K + \frac12\log p $$
This is a straight line equation of type $( y = c + mx)$ as given in the question. However, I think the question has made a mistake saying the slope is $2$, but instead it should be $\frac12$. Accordingly, the equation with given numeric values should be:
$$ \log \frac xm = \log 3 + \frac12\log p $$
Or without logarithm:
$$ \frac xm = 3p^\frac12 $$
On plugging $p = 4$, You'll get:
$$ \frac xm = 3 \times 4^\frac12 = 3 \times 2 = 6 $$
Hence, you get the given answer.