NIMCET 2008 — Mathematics PYQ

To solve this easily, we use the property ${\log }_b(a)=\frac{1}{{\log }_a(b)}$. This allows us to move the variable terms into the argument of the logarithm:

Step 2: Apply the product rule for logarithms

Recall that ${\log }_a(MN)={\log }_a(M)+{\log }_a(N)$. Applying this to the denominators:

• ${\log }_a(ax)={\log }_a(a)+{\log }_a(x)=1+{\log }_a(x)$

• ${\log }_a(a^{2}x)={\log }_a(a^2)+{\log }_a(x)=2+{\log }_a(x)$

Step 3: Simplify with substitution

Let $t={\log }_a(x)$. The equation becomes:

Step 4: Solve for $t$

Find a common denominator: $t(1+t)(2+t)$.

Equating the numerator to zero:

Step 5: Factor the quadratic equation

So, $t=-\frac{1}{2}$ or $t=-\frac{4}{3}$.

Step 6: Determine the values of $x$

• If ${\log }_a(x)=-\frac{1}{2}⟹x=a^{-1/2}=\frac{1}{\sqrt{a}}$

• If ${\log }_a(x)=-\frac{4}{3}⟹x=a^{-4/3}=\frac{1}{\sqrt[3]{a^4}}$

Both values are valid as long as they don't make the bases of the original logarithms equal to $1$ or $\le 0$. Since a > 0 and $a\ne 1$, these values of $x$ are positive and distinct.

Conclusion:

There are 2 values of $x$ that satisfy the equation.

Correct Option: (a)