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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.

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

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

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  • $\begingroup$ dEmigOd also has a very nice approach here: Simply manipulate the equation $ab=1$ to find the desired result. $\endgroup$ – Mason Apr 24 '18 at 16:16
  • $\begingroup$ That's really nice. However, is there a more general way to proceed with $\log_\frac{a}{b} \frac{c}{d}$ ? $\endgroup$ – Frank Underwood Apr 24 '18 at 16:17
  • $\begingroup$ You mean that the sum is $\sqrt{c}$? and the difference is $\sqrt{d}$? For the result above we really need them to be $4$ apart right? So we can say this generally when $(a+b)^2=c$ and $(a-b)^2=c-4$ but I am not sure I am understanding what you're asking... $\endgroup$ – Mason Apr 24 '18 at 16:23
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Check out $$ab = 1$$

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

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  • $\begingroup$ Thanks for the answer. While this does the job for this problem, is there a more general way to proceed with $\log_\frac{a}{b} \frac{c}{d}$ ? $\endgroup$ – Frank Underwood Apr 24 '18 at 16:18

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