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

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Answer 1 · 2016-05-31T22:11:27

#x=log_2(5+sqrt29)-1#

Since #a^-b=1/a^b#, the equation can be rewritten as

#2^x-1/2^x=5#

Multiply each term by #2^x#.

#2^x(2^x)-1/2^x(2^x)=5(2^x)#

To simplify #2^x(2^x)#, use the rule #a^b(a^c)=a^(b+c)#:

#2^(x+x)-1=5(2^x)#

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

Let #u=2^x#. Then, #2^(2x)=(2^x)^2=u^2# and #5(2^x)=5u#.

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

#u=(-(-5)+-sqrt(5^2-(4*-1*1)))/(2*1)=(5+-sqrt29)/2#

Recall that #u=2^x#.

#2^x=(5+sqrt29)/2#

#x=log_2((5+sqrt29)/2)#

Using #log_a(b/c)=log_a(b)-log_a(c)#:

#x=log_2(5+sqrt29)-log_2(2)#

#x=log_2(5+sqrt29)-1#

As for the negative version of #(5+-sqrt29)/2#, we see that

#2^x=(5-sqrt29)/2#

Has no solutions since #(5-sqrt29)/2<0# and #2^x>0# for all values of #x#.

Answer 2 · 2016-05-31T22:13:50

#x = log_e(1/2 (5 + sqrt[29]))/log_e(2)#

Taking #y = 2^x# we have

#y -1/y=5->y^2-5y-1=0#

Solving for #y#

#y = 1/2 (5 - sqrt[29])#, #y = 1/2 (5 + sqrt[29])#

but #1/2 (5 - sqrt[29]) < 0# and #y = 2^x > 0# Now solving for #x#

so

#x = log_e(1/2 (5 + sqrt[29]))/log_e(2)#