What is log_2 (1/4(2^(x+1)))

Question

can you please explain the steps on how to do this??

 

Thank you.

Answer 1

Using the properties of logs,

 

 

log₂(1 / 4) + log₂(2^{[x + 1]}) =

 

log₂(1) - log₂(4) + (x + 1)log₂(2) =

 

0     -     2      + (x + 1) * 1 =

 

-2 + x + 1 =

 

x - 1

 

 

 

      

Answer 2

log₂[(1/4)(2^x+1)] = log₂(1/4) + log₂(2^x+1)

 

                                  = log₂(2^-2) + log₂(2^x+1)     

 

                                  = -2 + (x+1)                  log_a(a^x) = x

 

                                  = x - 1