$k2^x+2^x=8$, find the possible values of $k$

Question

Find all the possible values of $k$ such that equation $$k2^x+2^x=8$$ has a single root. Find the root in the case.

Can anyone give some hints for me? I have no idea how to solve it.

Answer 1

Hint: Factor it as $$(k+1)2^x=8$$

and divide by $k+1$ to obtain

$$2^x=\frac{8}{k+1}$$

provided that $k\neq-1$. What can we say about above equation?

Answer 2

Following what wythagoras did in his answer, we arrive at $$2^x = \frac{8}{k+1}$$

Now, the function $2^x$ is strictly increasing and is always positive. (*) So, we must have $$\frac{8}{k+1} > 0$$ and $k > -1$.

(*) To see what I mean, look at the graph below: the graph is the function $2^x$, as you can see the function hits every value greater than 0. So, you can find $x$ as long as you are given a positive number $c$ to the equation $2^x = c$.

The technical term for such an occurrence is that the function $f: \mathbb{R} \to \mathbb{R^{+}}, f(x) = 2^x$ is injective, allowing it to have an inverse as long as your co-domain or range is positive.

Answer 3

$ k 2^x+2^x=8 $

$=>2^x(k+1)=2^3$

$=>(k+1)=2^{3-x}$

we know that the value of exponential is all ways >0

That is $2^{3-x}>0$

So,

$k+1=2^{3-x}>0$

$=>k+1>0 $

and $2^{3-x}>0$

$=>k>-1$

And $2^{3-x}>0$