Question

how will i evaluate log(base3/2) 16/81 ? anyone !! please help me

Answer 1

is it
\[\log_{\frac{ 3 }{ 2 }} \frac{ 16 }{ 81 } \]

Answer 2

So first off:
\[\log_{b}M = x\]is the same exact thing as:
\[b^{x} = M\]
So with your problem, you have:
\[\log_{3/2}\frac{16}{81}= x\]Meaning we can rewrite this as
\[(\frac{ 3 }{ 2 })^{x} = \frac{16}{81}\]
So we need to raise 3/2 to some power that gets us 16/81. So one way to visually help us out is another rewrite. (3/2)^x can be rewritten as:
\[\frac{ 3^{x} }{ 2^{x} }\]Now we see something doesnt match. Powers of 3 go 3 9 27 81 and powers of 2 go 2 4 8 16. So from the looks of it, the powers of our numerators matchthe denominator of 16/81 and the powers of the denominator match the numerator of 16/81. In order to flip our fraction so numerators match numerators, we need to have a negative exponent. Negative exponents basically flip numbers from numerator to denominator and vice versa. So, as you may have notice, 3 9 27 81, 3^4 gives us 81 and 2 4 8 16, 2^4 gives us 16. But since we need it upside down, we just say that instead of the correct exponent being 4, we need it to be -4. So the answer in all of this is -4 because:
\[(\frac{ 3 }{ 2 })^{-4} = \frac{16}{81}\], which also is the same as saying:
\[\log_{3/2}\frac{16}{81} = -4\]

Answer 3

if it is, then
\[(\frac{ 3 }{ 2 })^x = \frac{ 16 }{ 81 }\]
Therfore,
\[(\frac{ 2 }{ 3 })^{-x} = \frac{ 16 }{ 81 }\]
Hence,
\[(\frac{ 2 }{ 3 })^{-4} = \frac{ 16 }{ 81 }\]
So x = -4

Answer 4

oh tnx... i got it... can i ask another question?