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.
