Can you solve this simple equation using algebra? I doubt it!

Graph the two functions 9x and 3^x and look for where the two intercept.

Do not doubt. For y = 9 x and y = 3^x,

At x = 0, 3^x > 9 x such that 1 > 0;

At x = 1, 9 x > 3^x such that 9 > 3;

At x = 2, 9 x > 3^x such that 18 > 9;

At x = 3, 3^x = 9 x such that 27 = 27;

At x = 4, 3^x > 9 x such that 81 > 36;

d y/ d x = 9 and d y / d x = (Ln 3) 3^x respectively where (Ln 3) 3^x > 9 at x >= 3+.

Largest value presumed real x only.

There are two intersect points, one at x = 0.127869 and another one at x = 3 only. Therefore, the largest value of x is 3.

$9x={ 3 }^{ x }..........(A)\\ Let\quad "3"\quad is\quad answer\\ Checking.\\ Putting\quad x=3\quad in\quad (A)\\ 9(3)={ 3 }^{ \left( 3 \right) }\\ =>\boxed { 27=27 } \\ Satisfy$

Guess and check really. 2 and 1 don't work but 3 works because 9*3 is the same thing as 3^3

take log base both sides

Easy just use log (3) 3is the base

(9)(3)=27 (3)(3)(3)=27 so x=3 after 3 no other number can be the value of x

9x=3^x -> x * 3^2= 3^x-> x=3 ^(x-2)-> x=3

9x3=3x3x3=27.so x=3

9×3=3^3 27=27

9*3=27 and 3^3=27 hence x=3

9 = 3^3; 9x = 3^3*x; so question is 3^2 * x = 3^x, assume x = 3^y, then question becomes 3^(2+y) = 3^(3^y) == 3^(2+y) = 3^(3y); 2 + y = 3y, y = 1, and follows x = 3.

9x3 = 3³--> 9+9+9=3x3x3 --> 27 = 27 --> Thus, in the Naturals numbers three is the single answer...

9*3=3^3=27