Question

Log_x²+x(x²-4)=log_4x²-6(x²-4)

Answer 1

$log_{x^2+x}(x^2-4)=log_{4x^2-6}(x^2-4)$
ОДЗ:
x² + x > 0 ⇔ x(x + 1) > 0 ⇔ x∈(-∞;-1)∪(0;∞)
x² + x ≠ 1 ⇔ x² + x - 1 ≠ 0 ⇔ x ≠ (-1 ± √5)/2
x² - 4 > 0 ⇔ (x - 2)(x + 2) > 0 ⇔ x∈(-∞;-2)∪(2;∞)
4x² - 6 > 0 ⇔ x² - 6/4 > 0 ⇔ (x - √6 / 2)(x + √6 / 2) > 0 ⇔ x∈(-∞;-√6/2)∪(√6/2;∞)
4x² - 6 ≠ 1 ⇔ x ≠ ± √7 /2
Пересечение x∈(-∞;-2)∪(2;∞)

$log_{x^2+x}(x^2-4)-log_{4x^2-6}(x^2-4)=0\\\frac{lg(x^2-4)}{lg(x^2+x)}-\frac{lg(x^2-4)}{lg(4x^2-6)}=0\\\frac{lg(x^2-4)lg(4x^2-6)-lg(x^2-4)lg(x^2+x)}{lg(x^2+x)lg(4x^2-6)}=0\\lg(x^2-4) *(lg(4x^2-6) - lg(x^2+x))=0\\1) lg(x^2-4)=0\\x^2-4=1\\x=\pm \sqrt{5} \\2) lg(4x^2-6)-lg(x^2+x)=0\\ lg(4x^2-6)=lg(x^2+x)\\4x^2-6=x^2+x\\3x^2-x-6=0\\D=1+72=73\\x=\frac{1\pm \sqrt{73}}{6} \notin ODZ$
Ответ: $\pm \sqrt{5}$

Answer 2

To solve the given equation, we will use some logarithmic properties and algebraic manipulations. Let's go step by step:

Step 1: Combine the logarithms on both sides using the logarithmic property: log(a) + log(b) = log(a * b).

log_x² + x(x² - 4) = log_4(x² - 6(x² - 4))

Step 2: Rewrite the expression inside the logarithm on the right side using a common factor.

log_x² + x(x² - 4) = log_4(x² - 6x² + 24)

Step 3: Simplify the expression on the right side.

log_x² + x(x² - 4) = log_4(-5x² + 24)

Step 4: Now, we can use the logarithmic property log(a) - log(b) = log(a / b) to rewrite the equation.

log_x² + x(x² - 4) = log_4(-5x² + 24)

log_x² + x(x² - 4) - log_4(-5x² + 24) = 0

Step 5: Now, we need to eliminate the logarithms. We can do this by converting the equation to exponential form.

For the first logarithm: log_x² = y means x^y = x².

For the second logarithm: log_4(-5x² + 24) = y means 4^y = -5x² + 24.

Step 6: Replace the logarithms with the exponential expressions.

x² + x(x² - 4) - (4^y) = 0

Step 7: Simplify the equation.

x² + x³ - 4x - 4^y = 0

Step 8: Now, we need to solve this cubic equation for x. Unfortunately, the general cubic equation cannot be solved using simple algebraic manipulations. You would need numerical methods or calculator tools to find the approximate solutions for x.

If you have a specific value for y, you can substitute it into the equation and then use numerical methods or a calculator to find the corresponding values of x. Otherwise, without a specific value for y, we cannot find an exact solution for x.

So, the equation log_x² + x(x² - 4) = log_4(x² - 6(x² - 4)) cannot be fully solved using simple algebraic techniques.