Give answer in step by step order

Question

x^log₂x+4  =32

Answer 1

1) Take the log₂ of both sides:

 

log₂(x^log₂^{x +4} ) = log₂ 32  (=> 5)

 

Use log property: log_b(x^y) = y · log_b(x)

 

(log₂x + 4)(log₂x) = 5

 

(log₂x)² + 4(log₂x) - 5 = 0

 

Let y = log₂x 

 

y² + 4y - 5 = 0

 

(y + 5)(y - 1) = 0

 

y = -5 or y = 1  

 

log₂x = -5 => x = 1/32   log₂x = 1 => x = 0 (0 to any power is 0, so this is not a solution)

 

Check:

 

(1/32)^log₂^{(1/32) + 4}  = 32

 

(1/32)^-5+4  = 32

 

(1/32)^-1 = 32

 

32 = 32  √