2
$\begingroup$

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?

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

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.

I don't think such an equation can be solved algebraically in general, but luckily 2 works.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.