Question

find the product of all solutions to the equation

Answer 1

\[x(\sqrt{2011})=x ^{\log_{2011}x }\]

Answer 2

take log to base 2011 on both sides (i'll represent by log only)
we then have
log x + log(sqrt(2011)) = (log x)(log x)
=>log x + 1/2 = (log x)^2
its a quad eqn in log x
solve for log x and then x..
was that clear enough?

Answer 3

x~~5.89559 = (√(2011)/(log(2011))
x~~0

Answer 4

ok.. i understand... but i 4got to say it must solve without using calculator

Answer 5

hmmn,,leme try
we have
2(log x)^2 - 2logx -1 =0
log x = [2+ sqrt(12)]/2 = 1+ sqrt(3)
so x = 2011^(1+ sqrt3)
=> x = 2011 . 2011^sqrt(3)
i dont think you can do much here..theres only 1 soln
if you really dont wanna use calculator,,you may use calculus for approximations,,but using calculator would be the smarter thing.. ;)

Answer 6

if a quadratic equation has root \(r_1\) and \(r_2\) it can be written as:\[a(x-r_1)(x-r_2)=0\]if we expand this you see we get:\[a(x^2-(r_1+r_2)x+r_1r_2)=0\]\[ax^2-a(r_1+r_2)x+ar_1r_2=0\]which is usually written as:\[ax^2+bx+c=0\]therefore, the product of the roots can be obtained by dividing c by a

Answer 7

in your case this would give \(-\frac{1}{2}\)

Answer 8

@asnaseer we dont have to find products of solns of log x ..but we need them for x..

Answer 9

and x has only 1 real soln i'd guess..

Answer 10

ok, so we know that:\[\log_{2011}r_1\times\log_{2011}r_2=-\frac{1}{2}\]

Answer 11

hmmn,,but only one of them is real..other is complex..

Answer 12

i mean only 1 of em yields real value for x..

Answer 13

from what you have done above, we know:\[\log_{2011}x=\frac{1}{2}(1\pm\sqrt{3})\]therefore:\[x_1=2011^{\frac{1}{2}(1+\sqrt{3})}\]\[x_2=2011^{\frac{1}{2}(1-\sqrt{3})}\]therefore:\[x_1\times x_2=2011^{\frac{1}{2}(1+\sqrt{3})}\times 2011^{\frac{1}{2}(1-\sqrt{3})}=2011^{\frac{1}{2}(1+\sqrt{3}+1-\sqrt{3})}=2011\]

Answer 14

oh am so sorry @asnaseer ..never noticed that we should consider the other root too..
you did it perfectly! :)

Answer 15

don't worry @shubhamsrg - we all make mistakes sometimes - that is what makes us human :)

Answer 16

we solve until we get
\[\log_{x} (x \times \sqrt{2011})=\log_{2011}x \]\[\log_{x} x +\log_{x} \sqrt{2011}=\log_{2011}x\]\[1 +(1/2)\log_{x} 2011=\log_{2011}x\]\[1 +1/(2\log_{2011} x)=\log_{2011}x\]\[2(\log_{2011} x)+1=2(\log_{2011}x)^{2}\]\[0=2(\log_{2011}x)^{2}-2(\log_{2011} x)-1\]

Answer 17

yes neoh147 - that is what shubhamsrg did above (note that the base of your log should be 2011 and not x)

Answer 18

hmmn..

Answer 19

ok i get it d.. thx a lot... both of u... @asnaseer @shubhamsrg

Answer 20

i.e.:\[x(\sqrt{2011})=x ^{\log_{2011}x }\]\[\log_{2011}x(\sqrt{2011})=\log_{2011}x^{\log_{2011}x}\]\[\log_{2011}x+\log_{2011}2011^{0.5}=\log_{2011}x\times\log_{2011}x\]\[\log_{2011}x+0.5=(\log_{2011}x)^2\]\[2\log_{2011}x+1=2(\log_{2011}x)^2\]etc...

Answer 21

yw