Question

values of x satisfying the equation $10^\frac{2}{x}+25^\frac{1}{x}=4.25*50^\frac{1}{x}$

Answer 1

$\large\text{\underline{Basics}}$

$\red{\text{Laws of Indices}}$

$\purple{\bigstar\ (ab)^{n}=a^{n}b^{n}}$

$\implies10^{\frac{2}{x}}=2^{\frac{2}{x}}\times5^{\frac{2}{x}}$

$\implies50^{\frac{1}{x}}=2^{\frac{1}{x}}\times5^{\frac{2}{x}}$

$\bigstar\ \purple{(a^{m})^{n}=a^{mn}}$

$\implies 25^{\frac{1}{x}}=(5^{2})^{\frac{1}{x}}=5^{\frac{2}{x}}$

$\large\text{\underline{Solution}}$

$\implies2^{\frac{2}{x}}\times5^{\frac{2}{x}}+5^{\frac{2}{x}}=\dfrac{17}{4}\times2^{\frac{1}{x}}\times5^{\frac{2}{x}}$

Dividing both sides by $5^{\frac{2}{x}}$, we eliminate the common factor;

$\implies 2^{\frac{2}{x}}+1=\dfrac{17}{4}\times2^{\frac{1}{x}}$

Let's take substitution of $2^{\frac{1}{x}}=t$;

$\implies t^{2}+1=\dfrac{17}{4}t$

$\implies 4t^{2}+4=17t$

$\implies4t^{2}-17t+4=0$

However,

$\implies (4t-1)(t-4)=0$

$\implies t=\dfrac{1}{4}\text{ or }t=4$

Again, taking substitution of $2^{\frac{1}{x}}=t$;

$\implies2^{\frac{1}{x}}=2^{-2}\text{ or }2^{\frac{1}{x}}=2^{2}$

We compare the exponents now;

$\implies \dfrac{1}{x}=-2\text{ or }\dfrac{1}{x}=2$

$\implies x=-\dfrac{1}{2}\text{ or }x=\dfrac{1}{2}$

$\large\text{\underline{Conclusion}}$

The value that satisfies the equation is $x=-\dfrac{1}{2}\text{ or }x=\dfrac{1}{2}$.