We have:

#2^x-2^-x=5#

Take the #5# over to the other side to get:

#2^x-5-2^-x=0#

Now multiply the whole thing through by #2^x# and we get:

#2^x(2^x-5-2^-x)=(2^x)^2-5*2^x-1=0#

We substitute #t=x^2# and that gives us:

#t^2-5t-1=0#

We now have a quadratic that can be solved using the quadratic formula with terms #a=1, b=5# and #c=-1#.

#-> t = (5 +- sqrt(5^2-4*1*(-1)))/(2*1)=(5+-sqrt(29))/2#

Thus #t= 5/2+sqrt(29)/2# or #t=5/2-sqrt(29)/2#

Remeber #2^x=t#. Reverse the substitution of #t# to obtain:

#2^x=5/2+sqrt(29)/2#

Now take the logarithm to the base #2# and we get:

#x = log_2(5/2+sqrt(29)/2)~~2.37645#

We can ignore the #t=5/2-sqrt(29)/2# as it would involve taking the logarithm of a negative number which we wish to avoid.