Если x0 корень уравнения log2x+log4x+log16x=7 то выражение√x0+2 равно

log₂x + log₄x + log₁₆x = 7

log₂x + 1/2 log₂x + 1/4 log₂x = 7

4/4 log₂x + 2/4 log₂x + 1/4 log₂x = 7

7/4 log₂x = 7

log₂x = 7 * 4/7

log₂x = 4

x = 2⁴ = 16

x₀ = 16

√x₀ + 2 = √16 + 2 = 4+2 = 6

ответ: 6

Х+9=27        х-7=14         60-с=18        2х+1=77              4х-3=29
х=27-9         х=14+7        с=60-18        2х=77-1                4х=29+3  
х=18            х=21            с=42            2х=76                   4х=32 
18+9=27       21-7=14       60-42=18      х=76/2                  х=32/4   
                                                          х=38                     х=8        
                                                          2*38+1=77            4*8-3=29
                                 

Х+9=27        х-7=14         60-с=18        2х+1=77              4х-3=29
х=27-9         х=14+7        с=60-18        2х=77-1                4х=29+3  
х=18            х=21            с=42            2х=76                   4х=32 
18+9=27       21-7=14       60-42=18      х=76/2                  х=32/4   
                                                          х=38                     х=8        
                                                          2*38+1=77            4*8-3=29