# If `log_3 x + log_9 x + log_27 x = 4` , what is x.

justaguide | College Teacher | (Level 2) Distinguished Educator

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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)`

vaaruni | High School Teacher | (Level 1) Salutatorian

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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