Given $\log2=0.3010300$, $\log3=0.4771213$, $\log7=0.8450980$, find $\log0.3048$

Question

Given $\log2=0.3010300$, $\log3=0.4771213$, $\log7=0.8450980$, find $\log0.3048$

$\log0.3048=\log\left(\dfrac{3048}{10000}\right)=\log\left(\dfrac{2^3\cdot3\cdot127}{10^4}\right)$

$\Rightarrow 3\log2+\log3+\log127-4\log10$

Problem is I don't know what $\log127$ is. So there must be another approach but I don't see how else to factorize $0.3048$.

In base $10$, $\log 2=.301\cdots$. Is that what you meant to write? Your $\log 3$ is correct for base $10$ so I figure you have a typo. — lulu, Nov 29 '23 at 11:48
So then you mistyped $\log 2$. And note that $\log_{10} 127$ is close to $7\times \log_{10}2$ but it's not clear how precise an answer you were hoping for. — lulu, Nov 29 '23 at 11:50
That $3048$ looks like somebody is mixing $2048$ ($2 \times 2^{10}$) and $3072$ ($3 \times 2^{10}$) :-) — Dominique, Nov 29 '23 at 15:36

score 3 · Accepted Answer · answered Nov 29 '23 at 12:47

This is probably not how the problem is supposed to be solved, but a computer search yields $$\frac{2^{13} 7^{11}}{3^{12} 10^8} = \frac{63274455776}{207594140625} \approx 0.3047988521520\ldots, \tag{1}$$ hence $$\log 0.3048 \approx 13 \log 2 + 11 \log 7 - 12 \log 3 - 8 \approx -0.515987,$$ whereas the exact value is $-0.515985\ldots$. By contrast, approximating $127$ with $128 = 2^7$ yields $-0.512579$ which is substantially worse.

The choice of exponents in $(1)$ minimizes the absolute deviation for all integer exponents in the range $[-20,20]$.

I really liked the idea! $+1$. – Etemon Nov 29 '23 at 14:25 — Etemon, Nov 29 '23 at 14:25

score 2 · Answer 2 · answered Nov 29 '23 at 14:22

(All logarithms in this comment are in base 10.)

For a more accurate approximation, observe that $\log$ is linear on small intervals and approximate $\log 127$ as either $(\log 126 + \log 128)/2$, or as $(\log 125 + 2 \log 128)/3$.

Consider something similar - let's say you know $\log 2$ and $\log 3$ and want to approximate $\log 49$. You have $\log 48 = 4 \log 2 + \log 3 \approx 1.681241$ and $\log 50 = 2 - \log 2 = 1.69897$; then $\log 49 \approx 1.690106$. The true value is $2 \log 7 \approx 1.690196$.

score 2 · Answer 3 · answered Nov 29 '23 at 15:26

2

Let's consider some values near 127:

$$\log 125 = \log \frac{1000}{8} = 3 - 3\log 2 \approx 2.0969100$$ $$\log 126 = \log 2 + 2\log 3 + \log 7 \approx 2.1003706$$ $$\log 128 = 7\log 2 \approx 2.1072100$$

Interpolating these three points with a parabola gives:

$$\log x \approx -0.0000136333 x^2 + 0.00688257 x + 1.44961$$

From which $\log 127 \approx 2.1038049$. (The true value, rounded to 7 decimal places, is $2.1038037$.) This gives:

$$\log 0.3048 \approx -0.5159850$$

answered Nov 29 '23 at 15:26

Dan

14,978

1

For comparison: the values I gave above are $(\log 126) + (\log 128)/2 \approx 2.1037903$ and $(\log 125 + 2 \log 128)/3 = 2.1037767$, so this approximation is better. – Michael Lugo Nov 29 '23 at 17:17

Given $\log2=0.3010300$, $\log3=0.4771213$, $\log7=0.8450980$, find $\log0.3048$

3 Answers3