$\log_9 71$ or $\log_8 61$

I am trying to know which one is bigger :$$\log_9 71$$ or $$\log_8 61$$ how can i know without using a calculator ?

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This may not be very scientific, but in this case I would just say $9^2 = 81$ is "relatively further off" from $71$ than $8^2 = 64$ is from $61$. So $\log_9 71 < \log_8 61 < 2$. –  TMM Nov 2 '12 at 16:44
@ TMM - what's the break even point? Replacing $9$ by $a$, $71$ by $b$, $8$ by $c$ and $61$ by $d$, what's a nontrivial condition on $a$, $b$, $c$ and $d$ such that one will have inequality in one direction? –  Jonah Sinick Nov 2 '12 at 17:11
@Jonah: That's why I said it's not "very scientific" and why I used quotes for "relatively further off". It's more of a gut feeling ("the small difference between $a$ and $c$ does not compensate for the big difference between $b$ and $d$") than a careful analysis. –  TMM Nov 2 '12 at 17:25
@ern: Was this a question in an algebra-precalculus course? Because this is quite difficult, even with the use of calculus. –  TMM Nov 2 '12 at 21:24
@ TMM - I was just raising the question, wasn't criticizing your answer. –  Jonah Sinick Nov 3 '12 at 5:33

Ok, so we have that $$\frac{61}{64}>\frac{71}{81} \implies \log_{8}\left(\frac{61}{64}\right)>\log_{9}\left(\frac{71}{81}\right)\cdot\log_{8}(9)>\log_{9}\left(\frac{71}{81}\right)$$ By application of the change of base formula and the fact that $\log_{8}(9)>1$, which is trivial.

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slick solution! –  sourisse Nov 3 '12 at 0:31

$$\log_8 61 = \log_8\left(64\left(1-{3 \over 64}\right)\right) = 2 + \log_8\left(1 - {3 \over 64}\right)$$

$$\log_9 71 = \cdots = 2 + \log_9 \left( 1 - {10 \over 81}\right)$$

Let's drop the 2s, and note that both values are negative.

$$\log_8\left(1 - {3 \over 64}\right) = {\log(1 - 3/64) \over \log 8} = { 2 \log (1 - 3/64) \over 2 \log 8 } = {\log\left(\left(1-{3 \over 64}\right)^2\right) \over \log 64}$$

Now note that

$$\left(1-{3 \over 64}\right)^2 > 1 - {6 \over 64} > 1 - {6.4 \over 64} = 1 - {8.1 \over 81} > 1 - {10 \over 81}$$

so

$${\log\left(\left(1-{3 \over 64}\right)^2\right) \over \log 64} > {\log \left( 1 - {10 \over 81} \right) \over \log 64} > {\log \left( 1 - {10 \over 81} \right) \over \log 9} = \log_9 \left( 1 - {10 \over 81} \right)$$

Thus, $\log_8(61)$ is greater.

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How is $\frac{\cdots}{\log 64} > \frac{\cdots}{\log 9}$? –  Karolis Juodelė Nov 2 '12 at 22:35
@KarolisJuodelė - The numerator is negative. –  Rex Kerr Nov 2 '12 at 22:36

If we are allowed to use calculus, we can get a somewhat formal estimate of both numbers: $$\log_9 (71) = \log_9 (81) + \log_9(71/81) = 2 + \log_9 \left(1 - \frac{10}{81}\right) \approx 2 - \frac{1}{2 \ln 3}\left(\frac{10}{81}\right), \\ \log_8 (61) = \log_8 (64) + \log_8 (61/64) = 2 + \log_8 \left(1 - \frac{3}{64}\right) \approx 2 - \frac{1}{3 \ln 2}\left(\frac{3}{64}\right).$$ Using rough estimates like $\ln 2 \approx 0.7$, $\ln 3 \approx 1.1$, $\frac{10}{81} \approx \frac{1}{8} = 0.125$ and $\frac{3}{64} \approx \frac{3}{60} = 0.05$ we get $$\log_9 (71) \approx 2 - \frac{1}{2.1}(0.125) \approx 2 - 0.06 \approx 1.94, \qquad (\text{exact: } \log_9 (71) = 1.940\ldots) \\ \log_8(61) \approx 2 - \frac{1}{2.2}(0.05) \approx 2 - 0.025 \approx 1.975. \qquad (\text{exact: } \log_9(61) = 1.976\ldots)$$ So $\log_9 71 < \log_8 61$.

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You can change the base of the logarithm and put both of them in the same base and then you know that log whose base is bigger than 1 are crescent so u can easily find what's the biggest one.
Change formula: $\log_b x = \frac{\log_a x} {\log_a b}$

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How would you compute that without a calculator? –  ruakh Nov 2 '12 at 18:38
You can use the formula without a calculator and then it's easier to analize and extimate the values as also the other answers has shown,I just suggested to put the numbers in the same base to get things easier. –  Laura Nov 2 '12 at 20:54

I have a method which seems a bit too strange to be valid, but I thought I'd post it anyway to see what others thought. It is as follows:

Notice that $\log_8:(0,\infty) \to \mathbb{R}$ is strictly concave. This means that we have $$\log_8{\left( \frac{x+y}{2}\right)} > \frac{\log_8{x}+\log_8{y}}{2} \implies \log_8{\left( \frac{x+y}{2}\right)^2} > \log_8{(xy)}$$

Since $\log_8{9} >1$, we can conclude that $$\log_8{\left( \frac{x+y}{2}\right)^2} > \frac{\log_8{(xy)}}{ \log_8{9}}$$

Let $x = \sqrt{61}-i\sqrt{10}$ and $y = \sqrt{61}+i\sqrt{10}$. Plugging these in (this is okay because imaginary parts disappear), we find that $$\log_8{61}>\frac{\log_8{71}}{\log_8{9}} = \log_9{71}$$

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By drawing tangents of $\log$ curves at points $\log_9 81$ and $\log_8 61$ I obtain the following approximations: $$\log_9 71 < \log_9 81 - \frac{81-71}{81 \log 9} = 2 - \frac {10}{81 \log 9}$$ $$\log_8 61 > \log_8 64 - \frac{64-61}{61 \log 8} = 2 - \frac {3}{61 \log 8}$$ I can then show that $\frac{10}{81 \log 9} > \frac{3}{61 \log 8}$ as $\frac{10 \cdot 61}{81 \cdot 3} = \frac{610}{243} > 2 > \frac{ \log 9 }{\log 8}$

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