If log27 x + log9 x + log3x =11, then the value of x is

To solve the equation log27 x + log9 x + log3x = 11, we first need to understand that loga b = c means that a^c = b.

So, we can rewrite the equation as follows:

log27 x = log3^3 x = 3 log3 x log9 x = log3^2 x = 2 log3 x

Substitute these into the original equation:

3 log3 x + 2 log3 x + log3 x = 11 6 log3 x = 11 log3 x = 11/6

Now, we can rewrite this back into exponential form:

3^(11/6) = x

So, the value of x is 3^(11/6).