How do you solve #9^x+6^x = 4# ?

1 Answer
Jan 18, 2018

#x ~~ 0.346296#

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 #f(x) = 0#

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 #f(x)# is continuous, that means there must be a zero somewhere in #(0, 1)#. Linearly interpolating, a reasonable first approximation would be #a_0 = 0.15 ~~ 2/13#

Also since #f(x)# is strictly monotonically increasing with #x#, that real solution is unique.

#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# is an approximate zero of #f(x)# then a closer one is:

#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# we find:

#a_0 = 0.15#

#a_1 ~~ 0.391009#

#a_2 ~~ 0.348226#

#a_3 ~~ 0.346300#

#a_4 ~~ 0.346296#

#a_5 ~~ 0.346296#