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Kaylis [27]
3 years ago
15

Could someone please help me with this Math problem, Thank you

Mathematics
1 answer:
algol [13]3 years ago
3 0
11 = 2x - ( x'2 +1)
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Solve the following logarithmic equations.<br> log[(x^2 + 2x − 3)^4] = 0
saw5 [17]

Answer:

The solutions are x = 1.24 and x = -3.24

Step-by-step explanation:

Hi there!

First, let´s write the equation:

log[(x² + 2x -3)⁴] = 0

Apply the logarithm property: log(xᵃ) = a log(x)

4 log[(x² + 2x -3)⁴] = 0

Divide by 4 both sides

log(x² + 2x -3) = 0

if log(x² + 2x -3) = 0, then  x² + 2x -3 = 1 because only log 1 = 0

x² + 2x -3 = 1

Subtract 1 at both sides of the equation

x² + 2x -4 = 0

Using the quadratic formula let´s solve this quadratic equation:

a = 1

b = 2

c = -4

x = [-b± √(b² - 4ac)]/2a

x =  [-2 + √(4 - 4(-4)·1)]/2 = 1.24

and

x = [-2 - √(4 - 4(-4)·1)]/2 = -3.24

The solutions are x = 1.24 and x = -3.24

Have a nice day!

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