Logarithm Question

Question

If log₂sinx - log₂cosx -log₂(1-tan^2x) = -1,

Then find x.

Options :

A) (nπ)/2 + π/8, n€Z

B) nπ - π/8, n€Z

C) (nπ)/4  + π/2, n€Z

D) none of these

Answer is option A. But please tell me the method.

Answer 1

Notice that the left side of equation has common bases of logs.  So we use the properties of logarithms to write a single log expression on the left side, then equal that to -1.  But before we do so, we will need to rewrite the equation so that it will be easier to apply those properties.

 

log₂sinx - [log₂cosx + log₂(1 - tan²x)] = -1

 

 

Apply the properties of logs in the bracketed terms first.

 

log₂sinx - log₂(cosx(1 - tan²x)) = -1

 

 

Notice that we started with 3 log terms on the left side.  Now, we only have 2 log terms.  Apply the properties of logs once more to get only one log term.

 

log₂[sinx / (cosx(1 - tan²x))] = -1

 

 

The solution to a logarithm is the exponent to a log's base.

 

2^-1 = sinx / [cosx(1 - tan²x)]

 

 

Then notice that sin(x)/cos(x) is tan(x).

 

1/2 = tanx / (1 - tan²x)

 

 

Now solve for x from this equation.