Question

Rozwiąż równania

a)

b)

c)

d)

e)

f)

g) log przy podstawie 0,5 x do 2-log przy podstawie 0,5 z x= log przy podstawie 0,5 z 3

Answer 1

a)

2^(x^2 -x -3) = 0,5

2^(x^2 -x -3) = 2^(-1)

x^2 - x - 3 = -1

x^2 - x - 2 = 0

delta = 1 - 4*1*(-2) = 1 + 8 = 9

p(delty) = 3

x1 = [1 -3]/2 = -2/2 = -1

x2 = [1 +3]/2 = 4/2 = 2

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b)

(2^x)^2 - 3* 2^x + 2 = 0

Podstawiamy

y = 2^x

Mamy

y^2 - 3y + 2 = 0

delta = 9 -4*1*2 = 9 - 8 = 1

y1 = [3 -1]/2 = 2/2 = 1

y2 = [3 + 1]/2 = 4/2 = 2

zatem

2^x1 = 1  --->  x1 = 0

2^x2 = 2  --->  x2 =  1

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c)

log4 (log3 (log2 x)] = 0 <=> log3 (log2 x) = 4^0 = 1 <=>

<=> log2 x = 3^1 = 3 <=> x = 2^3 = 8

x = 8

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d)

log2 (x -1) = 3 <=> x-1 = 2^3 = 8 <=> x = 9

x = 9

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e)

log(x+5) + log(x-4) = log(x-3) + log(x +2)

log[(x+5)*(x-4)] = log[(x-3)*(x + 2)]

x^2 -4x +5x - 20 = x^2 +2x - 3x - 6

x - 20 = -x - 6

2x = 20 - 6

2x = 14

x = 7

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f)

log5 (x -6) = 3  <=> x -6 = 5^3 = 125 <=> x = 125 + 6 = 131

x = 131

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g)

log 0,5 (x^2) - log 0,5 (x) = log 0,5 (3)

log0,5 [(x^2)/ x] = log 0,5 (3)

log 0,5 [x ] = log 0,5 ( 3)

x = 3

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