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!

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- Nov 25th 2011, 04:41 PMbenny92000Solving for log
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! - Nov 25th 2011, 05:30 PMQuackyRe: Solving for log
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. - Nov 26th 2011, 06:17 AMbenny92000Re: Solving for log
I used the subtraction rule to get 1/2(y-x)