Question

Log7 (x+8)=1+log7 (3x-14)​

Answer 1

Решение:

Найдем ОДЗ:

$\begin{equation*}\begin{cases}x+8>0\\3x-14>0\end{cases}\end{equation*} \;\;\;\;\; \Rightarrow \;\;\; x> 4 \dfrac{2}{3}$

А далее - решим уравнение:

$log_7(x+8) = 1 + log_7 (3x-14)\\log_7 (x+8) = log_7 7 + log_7 (3x-14)\\log_7 (x+8) = log_7 (7 \cdot (3x-14))\\log_7 (x+8) = log_7 (21x-98)\\x+8=21x-98\\21x-x = 98+8\\20x=106\\x=5.3$

Заметим, что данный корень нам подходит (по ОДЗ).

Ответ: $\boxed { \rm x = 5.3.}$  

Answer 2

Ответ:

{5,3}

Пошаговое объяснение:

log₇ (x+8)=1+log₇ (3·x-14)​

ОДЗ:

$\displaystyle \left \{ {{x+8>0} \atop {3*x-14>0}} \right.$ ⇔ $\displaystyle \left \{ {{x>-8} \atop {x>\frac{14}{3} }} \right.$ ⇔ x∈(4 2/3; +∞)

log₇ (x+8) = log₇ 7 + log₇ (3·x-14)​

log₇ (x+8) = log₇ 7·(3·x-14)​

x+8 = 7·(3·x-14)

x+8 = 21·x-98

21·x - x = 8+98

20·x = 106

x = 106:20 = 5,3 ∈(4 2/3; +∞)

Answer 3

To solve the equation Log7(x+8) = 1 + Log7(3x-14), you can use the properties of logarithms. The goal is to isolate the variable x. Here's how you can do it step by step:

1. Start by subtracting Log7(3x-14) from both sides of the equation to get all the logarithmic terms on one side:

   Log7(x+8) - Log7(3x-14) = 1

2. Use the properties of logarithms to combine the two logarithmic terms on the left side. Specifically, you can use the quotient rule, which states that Log_b(a) - Log_b(c) = Log_b(a/c):

   Log7[(x+8)/(3x-14)] = 1

3. Now, you want to get rid of the logarithm. Rewrite the equation in exponential form:

   7^1 = (x+8)/(3x-14)

4. Simplify:

   7 = (x+8)/(3x-14)

5. To eliminate the fraction, multiply both sides of the equation by (3x-14):

   7(3x-14) = x + 8

6. Distribute the 7 on the left side:

   21x - 98 = x + 8

7. Now, isolate the variable x by moving all the terms with x to one side and constants to the other side:

   21x - x = 8 + 98

   20x = 106

8. Finally, divide both sides by 20 to solve for x:

   x = 106 / 20

   x = 5.3

So, the solution to the equation Log7(x+8) = 1 + Log7(3x-14) is x = 5.3.