Question

Proste równania wyładnicze

a)3^[x+2] x 9^[3x-1] = 27^[x+4] to w nawiasach kwadratowych znajduje się na miejscu potęgi

b) (0,5)^[x^2] x 2^[2x+2] = 1/64

Answer 1

mam nadzieję że pomogłem, liczę na naj:P

Answer 2

a)

3^[x+2] * 9^[3x - 1] = 27^[x + 4]

3^[x+2] * ( 3^2)^[3x - 1] = (3*3)^[x + 4]

3^[x+2] * 3^[6x - 2] = 3^[3x + 12]

3^[7x] = 3^[3x + 12]

7x = 3x + 12

4x = 12

x = 3

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b)

(0,5)^[x^2] * 2^[2x + 2] = 1/64

( 1/2)^[x^2]  * 2^[2x + 2] = 1/64

[ 2^(-1)]^[x^2] * 2^[2x + 2] = 1/64

2^[ -x^2] * 2^[2x + 2] = 1/64

2^[ -x^2 +2x + 2] = 2 ^[ - 6]

- x^2 + 2x + 2 = - 6

-x^2 + 2x + 8 = 0

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delta = 2^2 - 4*(-1)*8 = 4 + 32 = 36

p(delty) = 6

x = [ -2 - 6]/(-2) = -8/(-2) = 4

lub

x = [ -2 + 6]/(-2) = 4/(-2) = - 2

Odp. x = -2  lub x = 4

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