Question

Log2(3x-√x^2+6x)=1
Помогите решить. на фотографии подробно.

Answer 1

Log2( 3x - √(x^2 + 6x)) = 1
log2( 3x - √(x^2 + 6x)) = log2 (2)
3x - √(x^2 + 6x) = 2 
√(x^2 + 6x) = 3x - 2
x^2 + 6x = 9x^2 - 12x + 4
9x^2 - 12x + 4 - x^2 - 6x =  0
8x^2 - 18x + 4 = 0   /:2
4x^2 - 9x + 2 = 0 
D = 81 - 32 = 49
x1 = ( 9  + 7)/8 = 2
x2 = ( 9 - 7)/8 = 0,25

Проверка
log2( 6 - √(4 +12)) = 1
log2( 6 - 4) = 1
log2( 2) = 1

log2( - 0.5 ) ≠ 1
 

Ответ
2