Question

Log x+4 (x^2+2x-8)=1

Answer 1

Ответ:

Ответ указен на фотографии.

Объяснение:

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Answer 2

$log_{x + 4}( {x}^{2} + 2x - 8 ) = 1$

Составим систему:

{ х+4 = х²+2х-8

{ х+4>0

{ х²+2х-8>0

{ х+4≠1

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{ х²+х-12 = 0

{ х>-4

{ (x-2)(x+4)>0

{ х≠-3

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х²+2х-8 = 0

D = 2²-4*(-8) = 4+32 = 36 = 6²

x12 = (-2±6)/2

x1 = 2

x2 = -4

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x²+x-12 = 0

D = 1²-4*(-12) = 1+48 = 49 = 7²

x12 = (-1±7)/2

x1 = 3

x2 = -4 - не входит в решение

Ответ: х=3

Answer 3

To solve the equation $\log_{x+4}(x^2+2x-8) = 1$, we'll follow these steps:

Step 1: Rewrite the equation in exponential form: $x^2+2x-8 = (x+4)^1$

Step 2: Expand the right side: $x^2+2x-8 = x+4$

Step 3: Move all terms to one side to set the equation to zero: $x^2+2x-8 - x - 4 = 0$ $x^2 + x - 12 = 0$

Step 4: Factor the quadratic equation: $(x+4)(x-3) = 0$

Step 5: Solve for x by setting each factor equal to zero: $x+4 = 0 \quad \text{or} \quad x-3 = 0$

This gives us two possible solutions: $x = -4 \quad \text{or} \quad x = 3$

However, we need to check the validity of these solutions, since we have a logarithm with a base of $x+4$, and logarithms are only defined for positive bases.

For $x = -4$, the base of the logarithm ($x+4$) would be zero, which is not allowed. Therefore, $x = -4$ is not a valid solution.

For $x = 3$, the base of the logarithm ($x+4$) would be $7$, which is positive. Therefore, $x = 3$ is a valid solution.

So, the only valid solution is $x = 3$.