Question

1) 2)㏒₃(x²+7x-5)=1

Answer 1

Ответ:

343^{x-2}=49

(7^3)^{x-2}=7^2

7^{3x-6}=7^2

3x-6=2

3x=8

x=8/3

㏒₃(x²+7x-5)=1

x²+7x-5=3^1>0

x²+7x-5=3

x²+7x-8=0

1+7-8=0 - значит

x1=1

x2=-8

Answer 2

To solve the equation $2 \log_3(x^2 + 7x - 5) = 1$, we'll follow these steps:

1. Isolate the logarithmic term:

   Divide both sides by 2: $\log_3(x^2 + 7x - 5) = \frac{1}{2}$

2. Convert to exponential form:

   Recall that $\log_a(b) = c$ is equivalent to $a^c = b$. Applying this to our equation: $3^{\frac{1}{2}} = x^2 + 7x - 5$

   Simplifying the left side: $x^2 + 7x - 5 = \sqrt{3}$

3. Rearrange and solve the quadratic equation:

   Move $\sqrt{3}$ to the other side: $x^2 + 7x - \sqrt{3} - 5 = 0$

   This is a quadratic equation, so we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   In this case, $a = 1$, $b = 7$, and $c = -\sqrt{3} - 5$.

   $x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-\sqrt{3} - 5)}}{2(1)}$

   $x = \frac{-7 \pm \sqrt{49 + 4\sqrt{3} + 20}}{2}$

   $x = \frac{-7 \pm \sqrt{69 + 4\sqrt{3}}}{2}$

   So, the solutions are:

   $x_1 = \frac{-7 + \sqrt{69 + 4\sqrt{3}}}{2}$

   $x_2 = \frac{-7 - \sqrt{69 + 4\sqrt{3}}}{2}$

These are the solutions for the equation $2 \log_3(x^2 + 7x - 5) = 1$. If you want decimal approximations, you can calculate them using a calculator.