Problem:
In a Martian civilization, all logarithms whose bases are not specified are assumed to be base b, for some fixed b≥2. A Martian student writes down
3log(xlogx)loglog(x)(x)=56=54
and finds that this system of equations has a single real number solution x>1. Find b.
Solution:
We begin with the second equation:
loglogbxlogx=54
To eliminate the fraction, multiply both sides by loglogbx:
logx=54⋅loglogbx
This equation tells us that x is equal to (logbx)54:
x=(logbx)54
We now substitute this expression for x into the first equation, which involves x=x1/2:
3logb(x⋅logbx)=56
Substituting x=(logbx)27 from the earlier expression for x, we get:
3logb((logbx)27⋅logbx)=56
Combining the powers:
3logb((logbx)28)=56
Bringing the exponent outside:
logb((logbx)84)=56
Now we simplify this:
84⋅logblogbx=56
Divide both sides by 84:
logblogbx=8456=32
Now return to the expression x=(logbx)54. To express this in terms of b, note that logbx=y implies x=by, so:
by=y54
We are given that y=logbx, and from above we know logby=32. That means:
logb(logbx)=32
So:
logbx=b2/3
Now plug this into the expression for x:
x=(logbx)54=(b2/3)54=b36
But we also know from before that x=b36 and x=6108 from our earlier step (since logbx=36 implies x=b36 and also (logbx)54=3654=x), so: