Question

Log3(2x+3)+log19(2x+3)=2+log3(2x+3)

Answer 1

Пошаговое объяснение:

$x = 179$

Теперь ты з*****

Answer 2

To solve the given logarithmic equation, we'll need to apply logarithmic properties and solve for the variable $x$. Let's work through the steps:

The equation given is: $\log_3(2x+3) + \log_{19}(2x+3) = 2 + \log_3(2x+3)$

Step 1: Combine the logarithms on the left side using the logarithmic addition rule: $\log_a(b) + \log_a(c) = \log_a(b \cdot c)$ $\log_3(2x+3) + \log_{19}(2x+3) = \log_3((2x+3) \cdot (2x+3))$

Step 2: Simplify the right side of the equation: $\log_3((2x+3) \cdot (2x+3)) = 2 + \log_3(2x+3)$

Step 3: Subtract $\log_3(2x+3)$ from both sides of the equation to isolate the constant term: $\log_{19}(2x+3) = 2$

Step 4: Convert the equation to exponential form to solve for $2x+3$: $2x+3 = 19^2$

$2x+3 = 361$

Step 5: Subtract 3 from both sides to solve for $2x$: $2x = 361 - 3$

$2x = 358$

Step 6: Divide by 2 to solve for $x$: $x = \frac{358}{2}$

$x = 179$

So, the solution for the equation is $x = 179$.