Question

X^2*log 343(3-x)=log3(x^2-6x+9)​

Answer 1

Учитывая тот факт, что в левой части уравнения число 243, то решение Вы можете увидеть во вложении.

Удачи Вам и успехов)!

Answer 2

х²㏒₃⁵(3-х)=㏒₃(3-х)²

ОДЗ 3-х>0, т.е. х∈(-∞;3)

(1/5)*х²㏒₃(3-х)=2㏒₃I3-хI, с учетом ОДЗ

(㏒₃(3-х))*(х²/5-2)=0

х=±√10, в ОДЗ входит  -√10; или ㏒₃(3-х)=0⇒3⁰=3-х; х=2

Ответ 2; -√10

Answer 3

To solve the equation $x^2 \cdot \log_{343}(3-x) = \log_3(x^2 - 6x + 9)$, we'll need to simplify the logarithms and solve for $x$.

Let's start by simplifying the logarithms using some logarithmic properties:

1. Change of Base Formula: $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$
2. $\log_a(a) = 1$
3. $\log_a(a^n) = n$

Applying these properties to the equation:

$$x^2 \cdot \log_{343}(3-x) = \log_3(x^2 - 6x + 9)$$

1. $\log_{343}(3-x) = \frac{\log_3(3-x)}{\log_3(343)} = \frac{\log_3(3-x)}{7}$

Now the equation becomes:

$$x^2 \cdot \frac{\log_3(3-x)}{7} = \log_3(x^2 - 6x + 9)$$

Let's remove the fractions by multiplying both sides by 7:

$$x^2 \cdot \log_3(3-x) = 7 \cdot \log_3(x^2 - 6x + 9)$$

Since the bases of the logarithms are the same (both are 3), we can equate the exponents:

$$x^2 \cdot \log_3(3-x) = \log_3((x^2 - 6x + 9)^7)$$

Now, we can remove the logarithms and solve for $x$:

$$x^2 (3 - x) = (x^2 - 6x + 9)^7$$

Simplify both sides:

$$3x^2 - x^3 = (x^2 - 6x + 9)^7$$

This is a polynomial equation. You can expand the right side and attempt to solve for $x$, but it's going to be a lengthy process and might not lead to a simple solution. If you're looking for a numerical solution, you can use numerical methods or graphing calculators.

Keep in mind that polynomial equations of high degree might have multiple solutions, and some of them might be complex. If you're looking for a specific solution or range of solutions, numerical methods would be the way to go.