How I can solve this : $-2^x+3^{x/2}+1=0$ without using numerical ways?

Question

This equation : $-2^x+3^{x/2}+1=0$ is confusing me , However it has only one solution which it is an integer $ x=2$ , But i can't resolve it using clear way , I have used the varibale change $y= x/2$ in order to transforme it in equation of degree 2 but am regret that is not such polynomial for which solving it using standard Method , Any way without using numerical methods, I feel the only way work is to use Lambert Function ?

Answer 1

There is a general result for this sort of equations.

Given any $a_1, \ldots, a_n \ne 0$ and $b_1 > b_2 > \cdots > b_n > 0$. The LHS of equations of the form $$a_1 b_1^x + a_2 b_2^x + \cdots + a_n b_n^x = 0\tag{*1}$$ is known as a (generalized) Dirichlet polynomial. In $1883$, Laguerre has proved:

1. The number of real solutions of $(*1)$ is bounded by the number of sign changes in $a_k$.
2. The difference between the number of solutions and number of sign changes is an even number.

This is known as Generalized Descartes' rules of signs.

For the equation at hand, $(a_1,a_2,a_3) = (-1,1,1)$ and $(b_1,b_2,b_3) = (2,\sqrt{3},1)$. Since there is one sign change in $a_k$, it has exactly one real solution. Since you already figure out the solution at $x = 2$, this is all the solution it has.

For a modern introduction of this result, see

• G.J.O Jameson, Counting zeros of generalized polynomials: Descartes' rule of signs and Laguerre's extensions, (Math. Gazette 90, no. 518 (2006), 223-234). An online copy can be found here.

Answer 2

My Idea is to devide sides of this equality $-2^x+3^{x/2}+1=0$ by $2^x$ to get this form:

$1=(\frac{\sqrt{3}}{2})^{x}+(\frac{1}{2})^{x}$ i.e to $(\cos^{x} (\pi /6))+\sin ^{x}((\pi/6))$ this implies strongly $x=2$

Answer 3

The curves $y=2^x$ and $y=3^{x/2}+1$ meet at $x=2$ and swich the domination.

Prior to $x=2$ the curve $y=2^x$ is below $y=3^{x/2}+1$

and after $x=2$ it is the other way around.

We can compare the rate of changes to show it analytically.

Answer 4

Using the equivalent equation as of the hint of zeraoulia rafik, one observes that the equation $$ -1+\left(\frac{\sqrt3}2\right)^x+\left(\frac12\right)^x=0 $$ is constant or strictly monotonously falling in all terms on the left side, so that also the whole left side as their sum is strictly falling.. Which proves that there can only be one root, which was already found by inspection to be $x=2$.