Solving for log

**benny92000** · November 25th 2011, 04:41 PM

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!

**Quacky** · November 25th 2011, 05:30 PM

We have $\log_a(5)$ and $\log_a(7)$ and a target of $\log_a(1.4^{0.5})$

Seems easiest to work backwards.

$\log_a(1.4^{0.5})$

$=\frac{1}{2}\log_a(1.4)$ as $\log(x^n)=n\log{x}$

The task becomes getting $\log_a(1.4)$ in terms of $\log_a(5)$ and $\log_a(7)$

Best approach?

Well, $\log_a(1.4)=\log_a(\frac{2\times{7}}{10})$

$=\log_a(\frac{7}{5})$

Think about your log laws here.

**benny92000** · November 26th 2011, 06:17 AM

I used the subtraction rule to get 1/2(y-x)

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