Question

If the equation ${2^x} + {4^y} = {2^y} + {4^x}$ is solved for y in terms of x, where $x < 0$ , then the sum of the solution will be:
$1.\,\,x{\log _2}\left( {1 - {2^x}} \right)$
$2.\,\,x + {\log _2}\left( {1 - {2^x}} \right)$
$3.\,\,{\log _2}\left( {1 - {2^x}} \right)$
$4.\,\,x{\log _2}\left( {{2^x} + 1} \right)$

Answer 1

Hint: There are some properties of the logarithm that should be known to solve this problem like ${\log _y}y = 1$ which we are going to use in this question. In some questions, if a variable is in power and constant is in the base then we can store this value in some other variable.

Complete step-by-step answer:
In the given question, we have
${2^x} + {4^y} = {2^y} + {4^x}$
${2^x} + {2^{2y}} = {2^y} + {2^{2x}}$
Let, $a = {2^x}\,,\,\,b = {2^y}$
$a + {b^2} = b + {a^2}$
$a - b = {a^2} - {b^2}$
$a - b = (a - b)(a + b)$ (using the formula ${m^2} - {n^2} = (m + n)(m - n)$ )
$(a - b) - (a - b)(a + b) = 0$
Taking $(a - b)$ common,
$(a - b)\left[ {1 - (a + b)} \right] = 0$
Now, there are two cases:
Case: I $a - b = 0$
$a = b$
Putting the values of $a$ and $b$
${2^x} = {2^y}$
If the bases are equal, then power should also be equal.
$x = y$
Case: II
$a + b = 1$
Putting the values of $a$ and $b$
${2^x} + {2^y} = 1$
${2^y} = 1 - {2^x}$
Taking log in both sides with base $2$
${\log _2}{2^y} = {\log _2}\left( {1 - {2^x}} \right)$
Now, using property $\left( {{{\log }_a}{m^n} = n{{\log }_a}m} \right)$
$y{\log _2}2 = {\log _2}(1 - {2^x})$
Again, using property $\left( {{{\log }_m}m = 1} \right)$
$y = {\log _2}(1 - {2^x})$
Therefore, sum $ = \,x + {\log _2}(1 - {2^x})$
Hence, the sum of the solution of y in terms of x is $x + {\log _2}(1 - {2^x})$.
So, the correct answer is “Option B”.

Note: It is very important to remember all the properties of the logarithm. We must be careful of taking the base of the log while solving a question. It should be taken as per the requirement of the question. Don’t take natural log always. Algebraic identities are also important to remember.