Question

The domain of definition of the function $y\left( x \right)$ given by ${2^x} + {2^y} = 2$ is

Answer 1

We can convert the given equation of the function to the form of y equals to some functions on x by rearranging and applying logarithms. Then we can take the values inside the logarithm to be positive. Then we can find the values of x at which the term inside the log is positive and this interval will give the domain of the function.

Complete step by step solution:
We are given the equation of the function as ${2^x} + {2^y} = 2$. We can convert it into the form $y = f\left( x \right)$.
We have ${2^x} + {2^y} = 2$.
On rearranging, we get,
$\Rightarrow {2^y} = 2 - {2^x}$
Now we can take logarithm to the base 2 on both sides of the equation.
$\Rightarrow {\log _2}{2^y} = {\log _2}\left( {2 - {2^x}} \right)$
We know that, $\log {a^b} = b\log a$ . Using this relation, we get,
$\Rightarrow y{\log _2}2 = {\log _2}\left( {2 - {2^x}} \right)$
We know that ${\log _a}a = 1$ . On applying this relation, we get,
$\Rightarrow y = {\log _2}\left( {2 - {2^x}} \right)$
We know that logarithmic functions are defined for only positive values. So, the function is defined only for positive values inside the log. So, we can write,
$2 - {2^x} > 0$
When x becomes one, then the function $2 - {2^x}$ becomes zero. So, the given function is not defined at $x = 1$.
For values of x greater than 1, the value of $2 - {2^x}$ we become negative as $2 < {2^x}$. So, the given function is not defined for x greater than 1.
Therefore, the condition $2 - {2^x} > 0$ satisfies only when $x < 1$.
So, the given function $y = {\log _2}\left( {2 - {2^x}} \right)$ is defined only when $x < 1$.
As x can take values from negative infinity to 1, excluding the value 1, we can write the interval as,
$\left( { - \infty ,1} \right)$

The domain of definition of the function $y\left( x \right)$ given by ${2^x} + {2^y} = 2$ is $\left( { - \infty ,1} \right)$.

Note:
Domain of a function is the values of the variables at which the function is defined. We used the concept of logarithm to solve this problem. We use the logarithm to the base 2 because the equation has the power of 2 and we can use the properties of logarithm to simplify the function. We can also use the natural logarithm, but we will have a constant term also coming in the equation. We must know that the log function is defined only for positive values and not defined at 0.