Question

Log3^2 x - log 3 x >0

Answer 1

Ответ:

$log_3^2x-log_3x>0\\ODZ: x>0\\t^2-t>0\\t\in(-\infty;0)\cup(1;+\infty)\\log_3x1\\1)\\3>1 => 0 x>3\\=>x\in(0;1)\cup(3;+\infty)$

Answer 2

To solve the inequality log₃²x - log₃x > 0, we can use logarithmic properties to simplify it first.

Recall the properties of logarithms:

1. logₐ(b) - logₐ(c) = logₐ(b/c)
2. logₐ(b²) = 2 * logₐ(b)

Now, let's apply these properties to the given inequality:

log₃²x - log₃x > 0

Using property 2:

2 * log₃(x) - log₃(x) > 0

Now, combine the logarithms:

log₃(x) > 0

Now, we need to get rid of the logarithm to solve for x. Recall that the logarithm is the exponent to which the base (in this case, 3) is raised to get the result (x).

The inequality logₐ(b) > 0 is true if and only if b > 1.

So, we have:

x > 1

Therefore, the solution to the inequality is x > 1.