JAMIA 2021 — Mathematics PYQ

$4(x-1)={\log }_2(x-3)$

Domain:

x - 3 > 0 \implies x > 3

Analysis:

For x > 3:

$f(x)=4x-4$

$g(x)={\log }_2(x-3)$

At $x\to 3^{+}$:

$f(3)=4(3)-4=8$

$g(3)\to -\infty$

Since f(3) > g(3) and the slope of $f(x)$ is $4$, while the slope of $g(x)=\frac{1}{(x-3)\ln 2}$ decreases for x > 3:

$f^{\prime }(x)=4$

$g^{\prime }(x)=\frac{1}{(x-3)\ln 2}$

For x > 3.5, g'(x) < 1. Since the linear function starts higher and grows faster, the graphs never intersect.

Number of solutions = 0