Question

Log3x+log9x+3 log27 x < или = 3  

Answer 1

log 3 x + 1/2log 3 x + log 3 x <= 3

2,5log 3 x <=3

0<x<3^(6/5)

Answer 2

We can simplify the left-hand side of the inequality using the following logarithmic identities:

log(a) + log(b) = log(ab) log(a^n) = n log(a)

Applying these identities, we get:

log3x+log9x+3 log27 x = log(3x*9x) + log(27x^3) = log(27x^4)

Substituting this back into the inequality, we get:

log(27x^4) <= 3

Using the exponential form of logarithms, we can rewrite this as:

27x^4 <= 10^3

Simplifying the right-hand side, we get:

27x^4 <= 1000

Dividing both sides by 27, we get:

x^4 <= 1000/27

Taking the fourth root of both sides, we get:

x <= (1000/27)^(1/4)

Therefore, the solution to the inequality is:

x <= (1000/27)^(1/4)

Note that since logarithms are only defined for positive numbers, we must have x > 0. So the complete solution is:

0 < x <= (1000/27)^(1/4)