Question

Log2(x^2-1)=log1/3(1/27)

Answer 1

Ответ:

$x_1=3\\\\x_2=-3$

Пошаговое объяснение:

Преобразуем правую часть, воспользовавшись следующими свойствами логарифма:

$log_{a^p}(b)=\frac{1}{p}\cdot log_a(b) \\\\log_a(b^n)=n\cdot log_a(b)$

Тогда:

$log_{\frac{1}{3} }(\frac{1}{27} )=log_{3^{-1}}(3^{-3})=-\frac{1}{1} \cdot (-3)\cdot log_3(3)=3$

Исходное равенство:

$log_2(x^2-1)=3\\\\log_2(x^2-1)=log_2(8)\\\\x^2-1=8\\\\x^2=9\\\\\left[\begin{array}{c}x_1=3\\x_2=-3\end{array}\right$

Answer 2

To solve the equation log2(x^2 - 1) = log1/3(1/27), we can start by using the logarithmic property that states loga(b) = logc(b)/logc(a). Applying this property to the given equation, we can rewrite it as:

log(x^2 - 1) / log(2) = log(1/27) / log(1/3)

Next, we simplify the right side of the equation:

log(1/27) / log(1/3) = log(3^(-3)) / log(3^(-1)) = -3 log(3) / -log(3) = 3.

Therefore, the equation becomes:

log(x^2 - 1) / log(2) = 3.

To isolate log(x^2 - 1), we multiply both sides of the equation by log(2):

log(x^2 - 1) = 3 log(2).

Now, let's rewrite both sides of the equation in exponential form:

x^2 - 1 = 2^3.

Simplifying the right side:

x^2 - 1 = 8.

Now, add 1 to both sides of the equation:

x^2 = 8 + 1.

x^2 = 9.

Taking the square root of both sides:

x = ±√9.

Therefore, the solutions to the equation log2(x^2 - 1) = log1/3(1/27) are x = 3 and x = -3.