# What is number of roots of equation 4^x+2^x=272?

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### 3 Answers

Another method is to use the fact that `f(x)=4^x+2^x` is an increasing function. If you're not expected to know calculus, then you can probably assume this, maybe by appealing to the graph.

If you know calculus, then this isn't too difficult to prove using derivatives.

Once you know it's an increasing function, then there can be *at most* one point where `f(x)=272.` Again by using calculus (specifically the Intermediate Value Theorem), or by trying small numbers until one works, we can say that there is *exactly one* value of `x` for which `4^x+2^x=272.`

You should come up with the following substitution, such that:

`2^x = t`

Replacing the variable in equation, yields:

`t^2 + t = 272 => t^2 + t - 272 = 0`

Using quadratic formula, yields:

`t_(1,2) = (-1+-sqrt(1 + 1088))/2 => t_(1,2) = (-1+-33)/2`

`t_1 = (-1+33)/2 => t_1 = 16`

`t_2 = (-1-33)/2 => t_2 = -17`

You need to solve for x the equations `2^x = t_1` and `2^x = t_2` , such that:

`2^x = t_1 => 2^x = 16 => 2^x = 2^4 `

Equating the powers yields:

`x = 4`

`2^x = t_2 => 2^x = -17` invalid because `2^x > 0`

**Hence, evaluating the number of solutions to the given equation, yields one solution, **`x = 4.`

`4^x+2^x=272`

rewrite equation as

`(2^2)^x+2^x-272=0`

`(2^x)^2+2^x-272=0`

Let `2^x=y` then above equation reduces to

`y^2+y-272=0`

which is quadratic equation.So it has two roots which may be real /complex.

`y^2+17y-16y-272=0`

`(y+17)(y-16)=0`

`y=-17`

`or y=16`

i.e.

`2^x=-17` which is not possible for any value of x.

and

`2^x=16=2^4`

`x=4`

Thus it has one real root and other root not possible.