Can you Solve Cambridge University Admission Question ?

[@christopherward2748]

Very elegant way of doing it!

[@prollysine]

1/Vx=log((1+V5)/2)/log5 , x=(log5/log((1+V5)/2))^2 , x=~ 11.186027 ,

[@ManojkantSamal]

X= log 5/{log (3+*5)- log 2} or
X=log 5/{ log (3-*5)-log 2}..... May be
*= read as to the power
^=read as square root
As per question
25^(1/*x)+125^(1/*x)=625^(1/*x)
5^(2/*x)+5^(3/*x)=5^(4/*x)
Let 5^(1/*x)=R
So,
R^2+R^3=R^4
R^2(1+R)=R^4
1+R=R^4/R^2
1+R=R^2
R^2-R-1=0
Here a=1,b=-1, c=-1
D=b^2- 4ac
=(-1)^2-(4×1×-1)
=1+4=5
*D=*5
R=(-b±*D)/2a
={-(-1)±*5}/(2×1)
=(1±*5)/2
R^2=(1+5+2*.*5)/4 or
(1+5-2.*5)/4
R^2=(6+2.*5)/4 or (6-2.*5)/4
=2(3+*5)/4 or 2(3-*5)/4
=(3+*5)/2 or (3-*5)/2
Again
R^2={5^(1/*x)}^2
=[5^{(1/x)^(1/2)}^2
=5^(1/x)
So,
5^(1/x)=(3+*5)/2
Take the log
log {5^(1/x)}=log{(3+*5)/2}
(1/x). log 5 =log(3+*5) - log 2
(1/x)={log(3+*5)-log2 }/log 5
X=log 5/{log(3+*5) -log2 }
Similarly Taking the log we'll get
X=log 5/{log(3-*5) - log2 }

[@superacademy247]

Thanks for sharing your method 😎💯💕✅🙏💖