The sum of two numbers a and b is ${\sqrt 18}$ and their difference is ${\sqrt 14}$ . How do I find ${\log_ba}$?

Question

I've tried doing something like this

$${a + b + (a-b) = \sqrt 18 + \sqrt 14}$$ $${2a = \sqrt 18 + \sqrt 14}$$ $${2a = 3\sqrt 2 + \sqrt2\sqrt7}$$ $${2a = \sqrt 2( 3+\sqrt 7)}$$ $${a = \frac{(3+\sqrt 7)}{\sqrt 2}}$$

Similarly, I've got $$b = \frac{3-\sqrt 7}{\sqrt 2}$$

However, I have no idea how to go on from here. I have

$$\log_\frac{3-\sqrt 7}{\sqrt 2} \frac{(3+\sqrt 7)}{\sqrt 2}$$

How do I proceed from here? I know it must be very simple, but I can't seem to get it.

Answer 1

$$(a+b)^2=18$$ $$(a-b)^2=14$$ Expanding these equations and subtracting one from the other we find $4ab=4$. And therefore $ab=1$ and then $log_b$ both sides.

$$log_b(ab)=log_b(1)$$ $$log_b(ab)=0$$ $$log_b(a)+log_b(b)=0$$ $$log_b(a)+1=0$$ $$log_b(a)=-1$$

Answer 2

Check out $$ab = 1$$

So $a = \frac{1}{b}$.