Move all expressions to one side of the inequality:

#f(x)=12^x-5-6^(1-x)>0#

Taking the derivative of the left side with respect to #x# we have:

#f'(x)=log(12)12^x+log(6)6^(1-x)#

This expression is positive for all #x in RR#, therefore #f(x)# is strictly increasing at all times. This leaves two possibilities:

1) The expression is always positive and therefore the inequality is satisfied for all #x in RR#

2) The expression has a single root and satisfies the inequality for any #x# greater than that root.

If we can find any value of #f(x)# that is negative then the second possibility is true. We can see this is true for #f(0)#:

#f(0)=12^0-5-6^(1-0)=-10#

If we can solve #f(x)=0# then we solve this problem. There is no obvious way I can see to analytically solve this expression so I would use a numerical method. if you use the Newton-Raphson method (https://en.wikipedia.org/wiki/Newton%27s_method) with an initial guess of #x_0=0# then the series:

#x_(n+1)=x_n-f(x_n)/(f'(x_n))#

Will converge on the root. We find that this converges after a few iterations:

#x=0.756214 - #6 s.f.

Therefore we have the solution:

#x>0.756214# knowing this is a numerical approximation to the solution and only accurate to 6 s.f..