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Discriminant

The real root...

Question

The real roots of the equation $7^{l o g_{7} (x^{2} - 4 x + 5)} = x - 1$ are

1 and 2

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2 and 3

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3 and 4

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4 and 5

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Solution

The correct option is B 2 and 3
$a^{l o g_{a} x} = x, x^{2} - 4 x + 5 = x - 1 \Rightarrow x^{2} - 5 x + 6 = 0 \Rightarrow x = 2, 3$

Suggest Corrections

Similar questions

7^{{log}_{7} (x^{2} - 4 x + 5)} = (x - 1)

k

k x^{2} + k x + 1 = - 4 x^{2} - x

k = 5, - 3

x^{3} - 4 x^{2} + x + 6 = 0

1 :

2

1

2 :

1

2

x^{5} - 6 x^{2} - 4 x + 5 = 0

x^{4} - 4 x - 1 = 0

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