The equation `log_3 x + log_9 x + log_27 x = 4` has to be solved for x.

`log_3 x + log_9 x + log_27 x = 4`

Use the property: `log_a b = (log_x b)/(log_x a)`

=> `log_3 x + (log_3x)/(log_3 9) + (log_3 x)/(log_3 27) = 4`

=> `log_3x*(1 + 1/2 + 1/3) = 4`

=> `log_3x*(11/6) = 4`

=> `log_3x = 24/11`

=> `x = 3^(24/11)`

**The solution of the equation is **`x = 3^(24/11)`

Given: log3x + log9x + log27x = 4 , find x

Here we will change log9x and log27x to the base 3

Change of base rule :

Let we have given logax and

We require to change this to base n i.e. lognx

Then, lognx = (logax)/(logan) ------(1)

Applying the rule : log9x = (log3x)/(log39)

And log27x = (log3x)/(log327)

Substituting log9x and log27x to converted base 3 in

expression log3x + log9x + log27x = 4 we get,

=> log3x + (log3x)/(log39) + (log3x)/(log327) = 4

=> log3x + (log3x)/(log3(3^2)) + (log3x)/(log3 (3^3)) = 4

=> log3x + (log3x)/(2log33) + (log3x)/(3log33) = 4

=> log3x + (log3x)/2 + (log3x)/3 = 4

[ since log of any number to the same base = 1 ]

=> log3x + (1/2)(log3x) +(1/3) (log3x) = 4

=> log3x + (log3x(1/2)) + (log3x(1/3)) = 4 [using power law]

=> log3 x(1+1/2+1/3) = 4 [ using product law ]

=> log3 x(11/6) = 4

=> (11/6) log3x = 4

=> log3x = 4*(6/11) = 24/11

=> **x = 3(24/11) ****ß****Answer **

** **