Given equation : \((e^{2x} – 4)(6e^{2x} – 5e^x + 1) = 0\)
\(⇒ e^{2x} – 4 = 0 \;or \;6e^{2x} – 5e^x + 1 = 0\)
\(⇒ e^{2x} = 4 \;or\; 6(e^x)^2 – 3e^x – 2e^x + 1 = 0\)
\(⇒ 2x = ln4 \;or (3e^x – 1)(2e^x – 1) = 0\)
\(⇒ x = In2 \;or\; e^x = \frac{1}{3}\; or\; e^x = \frac{1}{2}\)
else, \(x = ln\frac{1}{3}, -ln2\)
Sum of all real roots = \(ln2 – ln3 – ln2\)
= \(–ln3\)
Hence, the correct option is (B): \(–log\;e^3\)
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 : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
The inverses of exponential functions are the logarithmic functions. The exponential function is y = ax and its inverse is x = ay. The logarithmic function y = logax is derived as the equivalent to the exponential equation x = ay. y = logax only under the following conditions: x = ay, (where, a > 0, and a≠1). In totality, it is called the logarithmic function with base a.
The domain of a logarithmic function is real numbers greater than 0, and the range is real numbers. The graph of y = logax is symmetrical to the graph of y = ax w.r.t. the line y = x. This relationship is true for any of the exponential functions and their inverse.
Exponential functions have the formation as:
f(x)=bx
where,
b = the base
x = the exponent (or power)