How do you solve #81^x = 243^x + 2#?

1 Answer
Mar 5, 2018

#"There is no real solution for the equation."#

Explanation:

#243 = 3*81#
#=> 81^x = (3*81)^x + 2#
#=> 81^x = 3^x * 81^x + 2#
#=> 81^x(1 - 3^x) = 2#
#=> (3^x)^4 (1 - 3^x) = 2#
#"Name "y = 3^x", then we have"#
#=> y^4 (1 - y) = 2#
#=> y^5 - y^4 + 2 = 0#
#"This quintic equation has the simple rational root "y= -1."#
#"So "(y+1)" is a factor, we divide it away :"#
#=> (y+1)(y^4-2 y^3+2 y^2-2 y+2) = 0#
#"It turns out that the remaining quartic equation has no real"# #"roots. So we have no solution as "y = 3^x > 0" so "y=-1#
#"does not yield a solution for "x.#

#"Another way to see that there is no real solution is :"#
#243^x >= 81^x" for positive "x", so "x" must be negative."#
#"Now put "x = -y" with "y" positive, then we have"#

#(1/243)^y + 2 = (1/81)^y#

#"but "0 <= (1/243)^y <= 1" and "0 <= (1/81)^y <= 1#
#"So " (1/243)^y + 2 " is always bigger than "(1/81)^y.#