Solve the equation ?

#4^2# + #2^(2x + 3)# = 40 + #4^(x+1)#

1 Answer
Sep 10, 2017

Real valued solution:

#x = (ln 6)/(2 ln 2)#

All solutions:

#x = (ln 6 + 2npii)/(2 ln 2)" "n in ZZ#

Explanation:

Given:

#4^2+2^(2x+3) = 40+4^(x+1)#

Note that #4=2^2#, so:

#4^(x+1) = (2^2)^(x+1) = 2^(2x+2)#

So we have:

#16+2^(2x+3) = 40+2^(2x+2)#

Subtract #16+2^(2x+2)# from both sides to get:

#2^(2x+3)-2^(2x+2) = 24#

That is:

#(2^3-2^2)2^(2x) = 24#

Dividing both sides by #2^3-2^2 = 2^2 = 4#, we get:

#2^(2x) = 6#

Taking natural logs of both sides:

#2x ln 2 = ln 6#

or for all solutions, including complex ones:

#2x ln 2 = ln 6 + 2npii" "n in ZZ#

Dividing both sides by #2 ln 2#, we get:

#x = (ln 6 + 2npii)/(2 ln 2)" "n in ZZ#