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^(x+1) = (2^2)^(x+1) = 2^(2x+2)#
So we have:
#16+2^(2x+3) = 40+2^(2x+2)#
Subtract
#2^(2x+3)-2^(2x+2) = 24#
That is:
#(2^3-2^2)2^(2x) = 24#
Dividing both sides by
#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
#x = (ln 6 + 2npii)/(2 ln 2)" "n in ZZ#