How do you solve #9^x+6^x = 4# ?
1 Answer
Explanation:
Given:
#9^x+6^x = 4#
Unfortunately this equation has no algebraic solution (even with the Lambert W function). We can find numerical approximations as follows.
Let:
#f(x) = 9^x+6^x-4#
We want to solve
Note that:
#f(0) = 9^0+6^0-4 = 1+1-4 = -2 < 0#
#f(1) = 9^1+6^1-4 = 9+6-4 = 11 > 0#
Since
Also since
#f'(x) = d/(dx) (9^x+6^x-4)#
#color(white)(f'(x)) = d/(dx) (e^(x ln9)+e^(x ln 6)-4)#
#color(white)(f'(x)) = e^(x ln9) ln 9+e^(x ln 6) ln 6#
#color(white)(f'(x)) = 9^x ln 9+6^x ln 6#
Newton's method tells us that if
#a_(i+1) = a_i - (f(a_i))/(f'(a_i))#
#color(white)(a_(i+1)) = a_i - (9^(a_i)+6^(a_i)-4)/(9^(a_i) ln 9 + 6^(a_i) ln 6)#
Applying this formula repeatedly, starting with
#a_0 = 0.15#
#a_1 ~~ 0.391009#
#a_2 ~~ 0.348226#
#a_3 ~~ 0.346300#
#a_4 ~~ 0.346296#
#a_5 ~~ 0.346296#