Question

Log1/7(3x-7)=log1/7(x+3)

Answer 1

Ответ:

5

Пошаговое объяснение:

$log_{\frac{1}{7} }(3x-7) = log_{\frac{1}{7} }(x+3)\\3x - 7 = x + 3\\2x = 10 \\x = 5$

Основания логарифмов одинаковые, поэтому мы можем их "Отбросить"

Answer 2

To solve the equation $\log_{\frac{1}{7}}(3x-7) = \log_{\frac{1}{7}}(x+3)$, we can use the fact that if $\log_a(b) = \log_a(c)$, then $b = c$.

So, in this case:

$3x - 7 = x + 3$

Now, let's solve for $x$:

$3x - x = 3 + 7$

$2x = 10$

$x = 5$

However, we need to check if this solution satisfies the original equation.

Plugging in $x = 5$ into the original equation:

$\log_{\frac{1}{7}}(3(5)-7) = \log_{\frac{1}{7}}(5+3)$

$\log_{\frac{1}{7}}(15-7) = \log_{\frac{1}{7}}(8)$

$\log_{\frac{1}{7}}(8) = \log_{\frac{1}{7}}(8)$

Since the logarithm bases are the same and the expressions inside the logarithms are equal, the solution $x = 5$ is valid.