Question

2) log_3(6 + 3 ^ x) = 3 - x

Answer 1

Ответ:

Ответ на фото

Пошаговое объяснение:

На фото, пошагово

Answer 2

Відповідь: х=1.

Покрокове пояснення:

Answer 3

To solve the equation $\log_3(6 + 3^x) = 3 - x$, we need to follow these steps:

Step 1: Rewrite the equation in exponential form. Step 2: Solve the resulting exponential equation. Step 3: Check for any extraneous solutions.

Let's start solving the equation step by step:

Step 1: Using the definition of logarithms, we have $3 - x = \log_3(6 + 3^x).$

Step 2: Now we can convert this into exponential form: $3 - x = \log_3(6 + 3^x)$ $3 - x = \log_3(3^1 + 3^x)$ $3 - x = \log_3(3 \cdot 3^{x-1})$ $3 - x = \log_3(3) + \log_3(3^{x-1})$ $3 - x = 1 + (x - 1)$ $3 - x = x$

Now we have a simple equation which can be solved for x.

$3 - x = x$ $3 = 2x$ $x = \frac{3}{2}$

Step 3: We need to check whether $x = \frac{3}{2}$ is a valid solution for the original equation, as sometimes the logarithmic equation might have extraneous solutions.

Let's substitute $x = \frac{3}{2}$ back into the original equation:

$\log_3(6 + 3^{\frac{3}{2}}) = 3 - \frac{3}{2}$

We can check whether both sides of the equation are equal. If they are, then $x = \frac{3}{2}$ is the solution. If they are not, then $x = \frac{3}{2}$ is not a valid solution.

Calculating the left-hand side:

$\log_3(6 + 3^{\frac{3}{2}}) = \log_3(6 + 3\sqrt{3})$

Calculate the right-hand side:

$3 - \frac{3}{2} = \frac{3}{2}$

It seems that the solution $x = \frac{3}{2}$ is not valid as both sides of the original equation do not match. Thus, the equation has no solution.