Question

If ${2^{\rm{x}}} - {2^{{\rm{x}} - 2}} = 192$, then the value of x is
(A) 5
(B) 6
(C) 7
(D) 8

Answer 1

Hint:
This question is based on an exponential function in which An exponential function is a Mathematical function in form f (x) = ${e^x}$, where “x” is a variable. The most commonly used exponential function base is the transcendental number e, which is approximately equal to 2.71828.

Complete step by step solution:
The above question is based on the exponential method which is An exponential function is a Mathematical function in form f (x) = ax, where “x” is a variable and “a” is a constant which is called the base of the function and it should be greater than 0. The most commonly used exponential function base is the transcendental number e, which is approximately equal to 2.71828.
The given expression ${2^{\rm{x}}} - {2^{{\rm{x}} - 2}} = 192$ can be simplified as follows!
Now,
According to the question;
 ${2^{\rm{x}}} - {2^{{\rm{x}} - 2}} = 192$
⇒ Take ${2^{\rm{x}}}$ as a common.
⇒ ${2^{\rm{x}}}\left( {1 - {2^{ - 2}}} \right) = 192$
⇒ ${2^{\rm{x}}}\left( {1 - \dfrac{1}{4}} \right) = 192$
⇒ ${2^{\rm{x}}}\left( {\dfrac{{4 - 1}}{4}} \right) = 192$
⇒ ${2^{\rm{x}}}\left( {\dfrac{3}{4}} \right) = 192$
⇒ ${2^{\rm{x}}} = \dfrac{{192 \times 4}}{3}$
After cancellation, we get,
⇒ $\dfrac{{{2^{\rm{x}}}}}{4} = \dfrac{{3 \times {2^6}}}{3}$
⇒ ${2^{\rm{x}}} \times 3 \times {2^{ - 2}} = 3 \times {2^6}$
⇒ ${2^{{\rm{x}} - 2}} = {2^6}$
⇒ Take power to we equal
⇒ x − 2 = 6
⇒ x = 8.

Note:
The above question is solved by two methods which are direct simplification method and other is the logarithmic method where the logarithmic method is the logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x.
we solve this problem using the logarithm
Taking log both sides, we get
 $\log {2^{{\rm{x}} - 2}}{\rm{\;}} = \log {2^6}$
⇒ x − 2 = 6
⇒ x = 8.