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write in logarithmic form (2/3)^-1 = 3/2
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on taking log both the sides you will get \[\log ((2/3)^ {-1})=\log (3/2)\] which gives you\[-1*\log (2/3)=\log(3/2)\]
\[\log a -\log b= \log (a/b)\]
can you do the rest?
log (3/2) + log (2/3) and Log a + log b = Log (a*b)
i think by "logarithmic form" they mean change \[b^x=y\] to \[log_b(y)=x\]
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so \((\frac{2}{3})^{-1}=\frac{3}{2}\) becomes \[\log_{\frac{2}{3}}(\frac{3}{2})=-1\]
but what i have done, is also one of the possibilities
uh... i stuck with -log(2/3)=log(3/2) because when i do evaluate that i get 1.
you will get zero log a + log b = log (a*b)= log ( 2/3 * 3/2) = log 1 =0
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