How do you solve #2^(x) - 2^(-x) = 5#?

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Answer 1 · 2016-07-08T17:56:06

#x = log_2(5+sqrt(29))-1 ~~ 2.3764522#

Let #t = 2^x#

Then the equation becomes:

#t-1/t = 5#

Multiply through by #4t# and rearrange slightly to get:

#0 = 4t^2-20t-4#

#= (2t-5)^2-29#

#= (2t-5)^2-(sqrt(29))^2#

#=((2t-5)-sqrt(29))((2t-5)+sqrt(29))#

#=(2t-5-sqrt(29))(2t-5+sqrt(29))#

Hence:

#2^(x+1) = 2*2^x = 2t = 5+-sqrt(29)#

For any Real value of #a#, #2^a > 0#.

So (for Real solutions) we require:

#2^(x+1) = 5+sqrt(29)#

Hence:

#x+1 = log_2(5+sqrt(29))#

So:

#x = log_2(5+sqrt(29))-1#

To calculate a numerical value for #log_2(5+sqrt(29))# we can use the change of base formula to find:

#log_2(5+sqrt(29)) = ln(5+sqrt(29))/ln(2) ~~ 3.3764522#

So:

#x ~~ 3.3764522 - 1 = 2.3764522#