Question

помогите ,пожалуйста!!!!!!№1логарифм Х по основанию 7+логарифм 36 по основанию 49=логарифм(2х+6)по основанию 1/7+логарифм 48 по основанию 7.№2х в степени логарифм х по основанию 3=1/9х в 3 степени.№38 в степени логарифм 5 по основанию 2-логарифм 3 по основанию 27.№4исследуйте функцию y=е (в степени 4х)(2-3х)на монотонностьи экстремумы

Answer 1

Решение первого примера есть у файле.

Answer 2

Task 1: Solving the equation involving logarithms

The equation you provided is:

log base 7 of x + log base 49 of 36 = log base (1/7) of (2x + 6) + log base 7 of 48

To solve this equation, we can use the properties of logarithms. Specifically, we can use the property that log base a of b + log base a of c = log base a of (b * c).

Applying this property to the equation, we can rewrite it as:

log base 7 of (x * 36) = log base (1/7) of (48 * (2x + 6))

Now, we can equate the expressions inside the logarithms:

x * 36 = 48 * (2x + 6)

Simplifying the equation:

36x = 48 * (2x + 6)

Expanding the right side:

36x = 96x + 288

Moving all the terms to one side:

36x - 96x = 288

Simplifying:

-60x = 288

Dividing both sides by -60:

x = -288/60

Simplifying the fraction:

x = -24/5

Therefore, the solution to the equation is x = -24/5.