Question

Log0,7(12x+1)>log0,7(6x+10)

Answer 1

Ответ:

( -1/12 ; 1,5 )

Пошаговое объяснение:

ОДЗ:

$12x + 1 > 0$

$x > \frac{ - 1}{12}$

$6x + 10 > 0$

$x > \frac{ - 5}{3}$

Неравенсво

$log_{0.7}(12x + 1) > log_{0.7}(6x + 10)$

$12x + 1 < 6x + 10$

$6x < 9$

$x < 1.5$

Answer 2

To solve the inequality log₀.₇(12x+1) > log₀.₇(6x+10), we need to apply some properties of logarithms. In particular, we can use the fact that if logₐ(b) > logₐ(c), then b > c, assuming a > 1.

Here's the step-by-step solution:

Step 1: Rewrite the inequality using the property mentioned above. 12x + 1 > 6x + 10

Step 2: Move all the x terms to one side and the constants to the other side of the inequality. 12x - 6x > 10 - 1 6x > 9

Step 3: Divide both sides by 6 to isolate x. x > 9/6 x > 3/2

So, the solution to the inequality is x > 3/2.