Problem: If log(base a)5 = x and log (base a)7 = y, then log (base a) 1.4^(1/2) =?
The answer is 1/2 x (y-x)
(I hope you follow my notation) Thanks!
We have $\displaystyle \log_a(5)$ and $\displaystyle \log_a(7)$ and a target of $\displaystyle \log_a(1.4^{0.5})$
Seems easiest to work backwards.
$\displaystyle \log_a(1.4^{0.5})$
$\displaystyle =\frac{1}{2}\log_a(1.4)$ as $\displaystyle \log(x^n)=n\log{x}$
The task becomes getting $\displaystyle \log_a(1.4)$ in terms of $\displaystyle \log_a(5)$ and $\displaystyle \log_a(7)$
Best approach?
Well, $\displaystyle \log_a(1.4)=\log_a(\frac{2\times{7}}{10})$
$\displaystyle =\log_a(\frac{7}{5})$
Think about your log laws here.