4^x+0,5=2^x+1 1-sinx=2 cos^2x lg (x+1) +lg (x-1) =lg8+lg (x-2) Segítsen valaki?!

4^x+0,5=2^x+1

(2^2)^x = 2^x + 0,5

2^2x - 2^x - 0,5 = 0

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

2*(2^x)^2 - 2*(2^x) - 1 = 0

(2^x - (1-√3)/2 )*(2^x - (1+√3)/2) = 0

2^x nem lehet (1-√3)/2 <0

ezert 2^x = (1+√3)/2

x = log_2 ((1+√3)/2 )

1-sin(x) = 2cos^2(x)

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

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

2sin^2(x) - sin(x) -1 = 0

(sin(x) -1)( 2sin(x) +1) = 0

Ket megoldas van sin(x)-re:

sin(x) = 1

ekkor x = Pi/2 + 2kPi

vagypedig

sin(x) = -1/2

x = 5Pi/4 + 2kPi

vagy

x = 7Pi/4 + 2kPi

lg(x+1) + lg(x-1) = lg(8) + lg(x-2)

x>2 kulonben nem lenne minden lg ertelmezheto.

lg(x+1)(x-1) = lg8(x-2)

(x+1)(x-1) = 8(x-2)

x^2 -1 = 8x -16

x^2 - 8x + 15 = 0

(x -3)(x-5) = 0

x=3

vagy

x = 5