1. ## log 3

1. If $log_{9}5=x$ and $log_{27}6=y$, find an expression in terms of $x$ and $y$ for $log_{3}1.2$.

2. Transform the logarithms in base 3:

$x=\frac{\log_3 5}{2}, \ y=\frac{\log_3 2+1}{3}$

$\log_3(1.2)=\log_3\frac{6}{5}=\log_3 2+1-\log_3 5=3y-2x$

3. I would have done this a slightly different way, using exponentials rather than logarithms. $log_9(5)= x$ so $9^x= 5$. $log_{27}(6)= y$ so $27^y= 6$.
$9= 3^2$ so $9^x= (3^2)^x= 3^{2x}= 5$ and $27= 3^3$ so $27^y= (3^3)^y= 3^{3y}= 6$.

Since 1.2= 6/5, we have $1.2= \frac{6}{5}$ $= \frac{3^{3y}}{3^{2x}}$ $= 3^{3y-2x}$ and so $log_3(1.2)= 3y- 2x$.