What rules are we using to show that $3^{-s}=\frac{1}{2}$ if $s=\frac{\log 2}{\log 3}$
I cannot understand how you can raise a number to a logarithm divided by a logarithm
What rules are we using to show that $3^{-s}=\frac{1}{2}$ if $s=\frac{\log 2}{\log 3}$
I cannot understand how you can raise a number to a logarithm divided by a logarithm
Applying the rule for change of bases, observe first that $\log_3 2 = \frac{\log 2}{\log 3}= s$, where the latter logarithms (in the ratio) can be to any base (but identical in numerator and denominator).
Then $3^{-s} = \frac{1}{3^s} = \frac{1}{2}$
because $3^{\log_3 2} = 2$