What's the value of log A at base 2?

Hi Leandro,

Multiply both sides by sin x and then repeatedly use the formula for sin 2x.

Chris

Leandro wrote back

I understand... I saw in one site this ident of trig:
sin (mx)(cos nx) = 1/2(sen (m+n)x + sen (m - n)x)

and A = cos x . cos 2x . cos 4x . cos 8x

Making sin x on the both sides:
sin x . A = sin x . cos x . cos 2x . cos 4x . cos 8x
If sin 2x = 2 sin x cos x, so sin x cos x = sinx 2x / 2. So
sin x . A = 1/2(sin 2x . cos 2x . cos 4x . cos 8x)
So, sin 2x . cos 2x = 1/2(sen 4x + sen 0x)
sin x . A = 1/4(sen 4x . cos 4x . cos 8x)
sin x . A = 1/8(sen 8x . cos 8x)
sin x . A = 1/16(sin 16x)
A = 1/16(sin 16x/ sin x)

But I don't know what to do anymore :(

Leandro,

Now you can use properties of the logarithm functions, in particular the log base 2 function.

First for any logarithm

log(a b) = log(a) + log(b) and log(a/b) = log(a) - log(b)

second, for the log base 2

log₂(2^p) = p

Thus

log₂(A)
= log₂(2^-4(sin 16x / sin x )
= log₂(2^-4) + log₂(sin 16x / sin x )
= -4 + log₂(sin 16x ) - log₂(sin x )

Penny

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