Animated Solution for Mathematics - Basic Mathematics: The number of solutions of the equation ${\log }_4(x-1)={\log }_2(x-3)$ is

Before we even touch the algebra, we must respect the laws of the logarithm. A logarithm ${\log }_a(b)$ is only defined when the argument $b$ is strictly greater than zero.

For our equation, this gives us two non-negotiable conditions: $x-1>0$ (which implies $x>1$) and $x-3>0$ (which implies $x>3$).

To satisfy both, we must exist in the intersection of these sets: $x>3$. This is our 'safe zone.' Any solution we find outside this region is a phantom, a mathematical mirage that we must reject.

Now, we face the base mismatch. We have base 4 on the left and base 2 on the right. We cannot equate arguments until the bases are identical.

We invoke the powerful property ${\log }_{a^n}b=\frac{1}{n}{\log }_{a}b$. Since $4=2^2$, we can rewrite the left side:

Now, our equation is harmonized:

To clear the fraction, we multiply both sides by 2, yielding ${\log }_2(x-1)=2{\log }_2(x-3)$. Next, we apply the power rule, $n{\log }_{a}b={\log }_a(b^n)$, to bring that coefficient of 2 inside the logarithm as an exponent:

With the bases now identical, we can safely equate the arguments: $x-1=(x-3{)}^2$. Expanding the right side gives us the quadratic:

Rearranging everything to one side, we arrive at the standard form:

Factoring this quadratic is a breeze: $(x-2)(x-5)=0$. This gives us two potential candidates: $x=2$ and $x=5$.

But wait! Remember our 'safe zone' from Phase 1? We established that $x$ must be greater than 3. The value $x=2$ fails this test miserably; it is an extraneous root.

Only $x=5$ survives the scrutiny of the domain. Thus, we have exactly one valid solution. The beauty of this problem lies not in the complexity of the calculation, but in the discipline of the verification.