Solve for X plz
\[x ^{\sqrt{x}}=\sqrt{x ^{x}}\]
x= 1 and x = 4
How i know the answer but how?
use logs
You could start by taking the natural log of both sides to make it easier.
Ahh i dont know much about logs
Can u show it plzz?
remember that \(ln(a^b)= b~ln(a)\)
yea
and also note: \(\large \sqrt{a} = a^\frac{1}{2}\)
Then?
can u show it step By step? PLzz
I know Identities. just show how u solved it
\(\large ln(x^{\sqrt{x}})= ln(\sqrt{x^x})\)
you know the rules. apply them. I don't want to simply give you the answer, I will help you step-by-step. But I also want you to try it first :) This is a good problem.
Man i am not able to do it plzz help i beg you? I have forgot logs i studied them a year before
Yes! You can do it! I am confident that you can do this! Trust me. I wouldn't be here if I didn't think you could do it.
I actually i studied logs just for fun i am in 9 thgrade and will learn logs in 12
i juyst remember identities nothing else
\(\sqrt{x}~ln(x) = ln(x^{x+\frac{1}{2}})\) does that make sense?
Yea
I added the \(x+\frac{1}{2}\) to make it easier. Since you can simply add exponents
Ohk.
Now, you try it from here. Add the \(x+\frac{1}{2}\) and move the exponent to the front of the ln like I did for the left side. REMEMBER: \(ln(a^n) = n~ln(a)\)
ACtually the problem is I am just able to do it till here after this i am stuck. I asked to solve it cause i thought i did this Wrong.
U dere?
Yes. I'm here.
Join our real-time social learning platform and learn together with your friends!