We start by dividing through by \( t^2 \) to simplify the equation:
\[ \frac{e^{2x}}{t^2} = t^4 - t^3 - 3t^2 - t + 1 = 0 \]
We make the substitution \( t = u \) and transform the equation further:
\[ t^2 + 1 = t^2 - t + 1 - 3 = 0 \] which simplifies to the quadratic equation: \[ u^2 - u - 5 = 0 \]
The quadratic equation \( u^2 - u - 5 = 0 \) has roots given by:
\[ u = \frac{1 \pm \sqrt{21}}{2} \]
Thus, the solutions for \( t \) are:
\[ t = 1 + \frac{\sqrt{21}}{2} \quad \text{or} \quad t = 1 - \frac{\sqrt{21}}{2} \]
There are two real values of \( t \), corresponding to the two roots of the transformed equation.
There are two real values of \( t \).
The domain of \(y= cos^{-1}|\frac{2-|x|}{4}| log(3 - x)^{-1}\) is [α, β) - {y} then the value of α+β-y =?
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)