How do I express #(2^x)-1=2/(2^x)# as a quadratic equation, and show that only one real solution, #x=1#, exists?

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Answer 1 · 2018-02-21T20:20:58

See explanation...

Given:

#2^x-1 = 2/(2^x)#

Multiply both sides by #2^x# to get:

#(2^x)^2-(2^x) = 2#

This is a quadratic equation in #2^x#. If you wish you can use a substitution #t = 2^x# to get #t^2-t=2#, but I will just leave it as #2^x#...

Subtract #2# from both sides to get:

#0 = (2^x)^2-(2^x)-2 = (2^x-2)(2^x+1)#

So:

#2^x = 2" "# or #" "2^x = -1#

Note that for any real value of #x# we have #2^x > 0#

Hence the only possible real solutions are given by:

#2^x = 2#

The function #f(x) = 2^x# is strictly monotonically increasing and therefore one to one.

We find:

#2^(color(red)(1)) = 2#

So #x=1# is the unique real solution.