Question

Log_x+1 (2x^2-5x-7)=2

Answer 1

${(x + 1)}^{2} = 2 {x}^{2} - 5x - 7 \\ {x}^{2} + 2x + 1 = 2 {x}^{2} - 5x - 7 \\ - {x}^{2} + 7x + 8 = 0 \\ x = - 1 \\ x = 8$

Answer 2

To solve the equation $\log_{x+1}(2x^2-5x-7) = 2$, we need to first rewrite the equation in exponential form. The logarithmic equation $\log_a(b) = c$ is equivalent to the exponential equation $a^c = b$.

In this case, we have: $\log_{x+1}(2x^2-5x-7) = 2$.

Rewriting in exponential form, we get: $(x+1)^2 = 2x^2 - 5x - 7$.

Now, let's solve this quadratic equation.

Step 1: Expand the left side of the equation: $x^2 + 2x + 1 = 2x^2 - 5x - 7$.

Step 2: Move all terms to one side of the equation to set it equal to zero: $2x^2 - x - 8 = 0$.

Step 3: We have a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 2$, $b = -1$, and $c = -8$. To solve this, we can use the quadratic formula:

The quadratic formula states that for an equation in the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step 4: Now, substitute the values of $a$, $b$, and $c$ into the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-8)}}{2 \cdot 2}$.

Step 5: Simplify the expression: $x = \frac{1 \pm \sqrt{1 + 64}}{4}$.

Step 6: Calculate the values: $x = \frac{1 \pm \sqrt{65}}{4}$.

Thus, the solutions to the equation are: $x = \frac{1 + \sqrt{65}}{4}$ and $x = \frac{1 - \sqrt{65}}{4}$.