Question

solve it please.....

Answer 1

y=x^xrootx

Log y=xrootx Log x

=x Log rootx

=x Logx +1/2 xLogx

=xrootx Log x=x Logx +1/2 xLogx

=x root x Log x-x Log x-1/2 xLog x=0

= (x root x - 3/2 x) Log x=0

x=0

Root x - 3/2=0

Root x =3/2

x=9/4

Log x =0

e^0=x

x=1

@skb

Answer 2

$Given:x^{x\sqrt{x}} = (x\sqrt{x})^x$

Apply 'log' on both sides, we get

$=> log(x^{x\sqrt{x}}) = log(x\sqrt{x})^x$

We know that log(aⁿ) = n log a.

$=>x\sqrt{x} log x=x log(x\sqrt{x})$

$=>x\sqrt{x} log x=xlogx^{1 + \frac{1}{2}}$

$=>x\sqrt{x} log x=x log x^{\frac{3}{2}}$

$=>x\sqrt{x} log x=\frac{3}{2}x log x$

$=>x\sqrt{x} log x=\frac{3x}{2}log x$

$=> x log x(\sqrt{x} - \frac{3}{2}) = 0$

(i)

x log x = 0

$=> log_{10}(x) = log_{10}(1)$

$=> x = 1$

(ii)

$=>\sqrt{x}=\frac{3}{2}$

$=>x=(\frac{3}{2})^2$

$=> x = \frac{9}{4}$

Therefore, the value of x = 1, 9/4.

Hope this helps!