Animated Solution for Mathematics - Basic Mathematics: The equation $x^{3/4({\log }_{2}x{)}^2+{\log }_{2}x-5/4}=\sqrt{2}$ has

The given equation is:

At first glance, this is intimidating because the variable $x$ appears in both the base and a complex quadratic expression in the exponent. This is the classic 'Variable Trap'.

In algebra, when you see a variable in the exponent, your instinct should immediately scream 'Logarithm!'. The logarithm is the only tool in our arsenal capable of pulling an exponent down to ground level.

Applying the Logarithm

We notice the exponent contains ${\log }_{2}x$. This is a massive hint. If we apply ${\log }_2$ to both sides of the equation, we can invoke the power rule: ${\log }_a(b^c)=c{\log }_a(b)$.

By taking ${\log }_2$ of both sides, the entire exponent jumps down to become a multiplier. The left side transforms from an exponential nightmare into a product:

On the right side, we have ${\log }_2(\sqrt{2})$. Since $\sqrt{2}=2^{1/2}$, this simplifies beautifully to $\frac{1}{2}$. We have successfully tamed the beast.

Substitution

Now, we have an equation that looks like this:

Writing ${\log }_{2}x$ repeatedly is not just tedious; it is a recipe for transcription errors. Let us introduce a bridge: the substitution $t={\log }_{2}x$.

Suddenly, the equation becomes a familiar algebraic friend:

By distributing the $t$ and multiplying the entire equation by $4$ to clear the fractions, we arrive at a clean, elegant cubic equation:

We are now in the realm of polynomials. To solve $3t^3+4t^2-5t-2=0$, we start with the method of inspection. Testing small integers is a classic JEE strategy.

If we test $t=1$, we get $3(1{)}^3+4(1{)}^2-5(1)-2=3+4-5-2=0$. Success! Since $t=1$ is a root, $(t-1)$ must be a factor.

Through polynomial division, we factor the cubic into $(t-1)(3t^2+7t+2)=0$. Further factoring the quadratic gives us:

The roots for $t$ are $1,-2,$ and $-\frac{1}{3}$.

We are almost home. We have the values for $t$, but the question asks for $x$. We must reverse our substitution using $x=2^t$.

For $t=1$, $x=2^1=2$.

For $t=-2$, $x=2^{-2}=\frac{1}{4}$.

For $t=-\frac{1}{3}$, $x=2^{-1/3}=\frac{1}{\sqrt[3]{2}}$.

We have found three distinct, real solutions: $x\in \{2,\frac{1}{4},\frac{1}{\sqrt[3]{2}}\}$. This problem is a perfect example of how a terrifying exponential equation can be dismantled, piece by piece, into a solvable polynomial.