Is it a Power Function or a Function in a Power?

Algebra Level 3

$\large 2^x + 2^{f(x)} = 2$

Consider a real valued function satisfying the equation above. Find the range of the function $f(x)$ .

Note: The domain of $f(x)$ is the largest subset of $\mathbb{R}$ allowed in which there exists a solution to the above equation.

(-\infty , 0)

(-\infty , 1)

None of these choices

(-\infty , -1)\cup(1,\infty)

(-\infty ,\infty)

(0,\infty)

(1,\infty)

1 solution

Sravanth C.
May 27, 2016

Rearranging the expression we get; $\begin{aligned} 2^x&=2-2^{f(x)}\\ \log_22^x&=\log_2(2-2^{f(x)})\\ x&=\log_2(2-2^{f(x)}) \end{aligned}$

Now if the RHS of the above equation should exsist then; $2-2^{f(x)}>0$ since the logarithm function is defined on positive integers greater than zero. Thus solving the inequality; $\begin{aligned} 2-2^{f(x)}&>0\\ 2^{f(x)}&<2^1\\ f(x)&<1\\ \end{aligned}$

Hence $f(x)\in (-\infty , 1)$ .

Note: This result can also be generalised, if we have; $a^x+a^{f(x)}=a; \quad\text{then } f(x)\in(-\infty ,1)\quad\quad\forall a>1$

Moderator note:

You have to be careful with the justification of each of these steps. What you have is a necessary condition, but you haven't shown that it is sufficient.

IE You should explain that, given any $y \in ( - \infty, 1 )$ , how do we find the value of $x$ to set $f(x) = y$ which shows that $y$ is in the range of $f(x)$ .

This is what lead to my "Note" in your problem.

Same solution here (+1) :)

Aditya Sky - 5 years ago

Is it a Power Function or a Function in a Power?

1 solution

Moderator note:

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