We have: #4^(2 x) - 17 (4^(x)) + 16 = 0#
#Rightarrow (4^(x))^(2) - 17 (4^(x)) + 16 = 0#
If you haven't noticed already, this equation is now in the form of a quadratic equation. #4^(x)# can be considered as a variable in this case.
Let's apply the quadratic formula:
#Rightarrow 4^(x) = frac(17 pm sqrt((- 17)^(2) - 4(1)(16)))(2(1))#
#Rightarrow 4^(x) = frac(17 pm sqrt(289 - 64))(2)#
#Rightarrow 4^(x) = frac(17 pm sqrt(225))(2)#
#Rightarrow 4^(x) = frac(17 pm 15)(2)#
#Rightarrow 4^(x) = 1, 16#
Let's solve for #x# for each case:
#Rightarrow 4^(x) = 1 and 4^(x) = 16#
#Rightarrow 4^(x) = 4^(0) and 4^(x) = 4^(2)#
Using the laws of exponents:
#therefore x = 0 and x = 2#
Therefore, the solutions to the equation are #x = 0# and #x = 2#.
