$\log_9 71$ or $\log_8 61$

Question

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

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 02 '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 02 '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 02 '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 02 '12 at 21:24
@ TMM - I was just raising the question, wasn't criticizing your answer. — Jonah Sinick, Nov 03 '12 at 05:33

score 16 · Answer 1 · answered Nov 03 '12 at 00:23

16

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.

answered Nov 03 '12 at 00:23

preferred_anon

17,765

3

slick solution! – sourisse Nov 03 '12 at 00:31

score 4 · Answer 2 · answered Nov 02 '12 at 22:15

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

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

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.

How is $\frac{\cdots}{\log 64} > \frac{\cdots}{\log 9}$? – Karolis Juodelė Nov 02 '12 at 22:35 — Karolis Juodelė, Nov 02 '12 at 22:35
@KarolisJuodelė - The numerator is negative. – Rex Kerr Nov 02 '12 at 22:36 — Rex Kerr, Nov 02 '12 at 22:36

sourisse · Answer 3 · 2015-07-31T03:07:07.407

4

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 $$\log_8{61}>\frac{\log_8{71}}{\log_8{9}} = \log_9{71}$$

edited Jul 31 '15 at 03:07

answered Nov 02 '12 at 23:52

sourisse

1,041

Cool! Writing 61 as a difference of the "squares" 71 and -10. BTW, the first inequality is just the logarithm of $\frac{x + y}{2} \ge \sqrt{xy}$ (and indeed one way to prove AMGM is via convexity of the logarithm). – Unit Sep 10 '19 at 14:27

score 3 · Answer 4 · edited Nov 02 '12 at 18:27

3

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: $\log_b x = \frac{\log_a x} {\log_a b}$

edited Nov 02 '12 at 18:27

Martin Sleziak

56,060

answered Nov 02 '12 at 17:09

Laura

761
5
15

How would you compute that without a calculator? – ruakh Nov 02 '12 at 18:38
1

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 02 '12 at 20:54

TMM · Answer 5 · 2012-11-02T17:31:19.567

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 $\log_9 71 < \log_8 61$.

score 1 · Answer 6 · answered Nov 02 '12 at 22:31

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 $\frac{10 \cdot 61}{81 \cdot 3} = \frac{610}{243} > 2 > \frac{ \log 9 }{\log 8}$

$\log_9 71$ or $\log_8 61$

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