This is not homework, it's just a brain teaser which I can't solve, just some hints should be sufficient, I know that from this we get: $$ (1/4)\log_2(p+q) = (1/2)\log_3 p = \frac{\log_3 q}{1+2\log_3 2} $$ now I'd like to combine these quantities in some way so I can see the value of $q/p$ from this, but I can't seem to be able to figure it out.
Asked
Active
Viewed 8,681 times
2
-
Does $,^{3}\log(n) = \log_{3}(n)$ in your question? Where $\log_{3}(n)$ is log base 3? – Thomas Russell Apr 22 '15 at 20:36
-
When you write \text{log}p you see $\text{log}p$ with no space between $\log$ and $p$, but when you write \log p, then you see $\log p$. That is standard usage. ${}\qquad{}$ – Michael Hardy Apr 22 '15 at 20:37
-
yes, I'm sorry for the confusion in notation, I'll change it – Holymonk Apr 22 '15 at 20:37
-
- thanks, you're right that's very pretty :)!
– Holymonk Apr 22 '15 at 20:42
2 Answers
5
So there is a number $x$ such that $9^{x}=p$, $12^x=q$ and $16^x=p+q$
Therefore $9^x + 12^x = 16^x$
One final hint: $12=3\times 4$
preferred_anon
- 17,121
-
jej, got it if the answer is (1/2)(1+\sqrt{5})! Thanks mate, I didn't think of using the definition of the logarithm I was just using the rules like $\log ab = \log a + \log b$! Superb hint! – Holymonk Apr 22 '15 at 21:05
-
No problem! I suppose the thing that tipped me off to use something different was the appearance of $p+q$ - logs don't like sums, so I didn't expect the log rules to be especially useful. Thanks! – preferred_anon Apr 22 '15 at 21:11
3
Let $L$ denote the common logarithmic value. Thus $p=9^L$, $q=12^L$, and $p+q=16^L$. Let $r=q/p=12^L/9^L=(4/3)^L$. Then
$$r^2=(16/9)^L=(p+q)/p=1+r$$
Can you take it from there?
Barry Cipra
- 79,832
-
Jep, thanks! same response as my response to Daniel Littlewood! :) I'd like to accept both your answers, but since I can only pick one I'm gonna have to go for the other 1 cause he was first, sorry! – Holymonk Apr 22 '15 at 21:06