can it be done??

Hey, is k constant for all x? does the value of k change according the values of x? Either way, I guess trial and error is the only way out ;)
 
i think k varies as x varies...he shud hav rather said k is some number.
for eg 2^2 = 4.. so k is 4 here , simlarly for 3 it it 27...
 
NeXgeN said:
k cannot be constant ....when x changes.... just take squareroot :P ...if u r using some special no.s then i dont know :)
Click to expand...

no
eg 4^4 = 256
if u take sqaure root , u get something else :P

@rocky , y do u say that? i hav attended all my classes :P
 
Last edited by a moderator:
Clearly u guys do not know maths......
what he means is
x^x = k
k is a constant.

So SOLVE and get the value of x for which the equation is true.
Unfortunately, I have forgotten how all this stuff is solved (forgot all that 1st and 2nd year engg maths) but IIRC there is no general purpose method to solve higher order equations.
You can do so, however, using a numerical method.
 
tifôsi said:
no
eg 4^4 = 256
if u take sqaure root , u get something else :P
Click to expand...
oh yeah...somehow i thought ^ as * ...my mistake.

solution is as bosky suggested take log on both sides
xlogx = log k

but again k cannot be constant if x varies.
 
ok...how does taking log on both sides solve the problem... it just restates it. It might be one of the steps, but def not the soln.
 
k cannot be constant.....as far as i can crack it......and anyway as Grease Says, Thaking Square just puts the same thing again in different words....does Not Solve it......
 
Afaik this is solved by iterative method , Chk out this prog. ,

void main()
{
double k="any number u wanna give",y=3,x=2;
while(fabs(x-y)>0.00001)
{ x=y;
y=log(k)/log(x);
}
cout<<x<<endl;
}

Edit: If u haave a scientific calculator this can be de pretty easily.
 
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