# Solving exponential equation

Here is the question:Solve $5^{\frac{x}{2}}-2^x=1$

How i tried:I was just looking at the equation and was trying different values of x and got x=2 .But the way to reach answer was not promising so I decided to graph it and observed that the function is ever increasing from (-$\infty,\infty$) ,so the graph cuts $y=1$ only once at $x=2$.This was my way to solve the question but is there some other algebric way to solve it?

• i have found $x=2$ – Dr. Sonnhard Graubner Jul 18 '15 at 9:57
• I have also found x=2 ,but what was your way to found x=2 @Dr.SonnhardGraubner – Kartik Watwani Jul 18 '15 at 9:58
• yes i have plot the graph of $f(x)=5^{x/2}-2^x-1$ – Dr. Sonnhard Graubner Jul 18 '15 at 10:00
• now you can differentiate $f(x)$ with respect to $x$ – Dr. Sonnhard Graubner Jul 18 '15 at 10:01
• Rewriting it like this, it is easy to see the solution $x = 2$: $$1 = 5^{x/2} - 2^x =5^{x/2} - 4^{x/2}$$ – izœc Jul 18 '15 at 10:01

The function $f(x)=5^{x/2}-2^x$ isn't increasing (It is decreasing to the left of Dr. SG's critical point). But it is less than $0$ when $x<0$.
So we have that $5^{x/2}-2^x<0$ when $x<0$ (so it can't equal $1$), and $5^{x/2}-2^x$ is increasing when $x>0$, with $f(0)=0$ and $\lim_{x\rightarrow\infty} f(x)=\infty$ so $5^{x/2}-2^x=1$ has exactly one solution.