MHB Find Leftmost Digit of $12^{37}$ Given Logs

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Find the leftmost digit of the figure $12^{37}$ given $0.3010<\log 2<0.3011$ and $0.4771<\log 3<0.4772$.
 
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We begin with:
$$12 = 2^2 \cdot 3$$
Using basic log properties with the information given:
$$22.2740 < \log(2^{74}) < 22.2814$$
$$17.6527 < \log(3^{37}) < 17.6564$$
Hence:
$$39.9267 < \log(12^{37}) < 39.9378$$
Now take $\delta = 40 - 39.9267 = 0.0733$ so that:
$$40 < \log(12^{37}) + \delta < 40.0111$$
Then the first digit of $10^{\log(12^{37}) + \delta}$ must be a 1 (and it must be just above a power of 10) and also:
$$10^{\delta} \approx 1.18$$
Note that:
$$10^{n} / 1 = 10^{n}$$
$$10^{n} / 1.111\cdots = 9 \cdot 10^{n - 1}$$
$$10^{n} / 1.25 = 8 \cdot 10^{n - 1}$$
And 1.18 falls comfortably between 1.11111... and 1.25, so [handwavy error analysis goes here] the first digit of:
$$\frac{10^{\log(12^37) + \delta}}{10^{\delta}} = 10^{\log(12^{37})} = 12^{37}$$
must be an 8.
 
Bacterius said:
We begin with:
$$12 = 2^2 \cdot 3$$
Using basic log properties with the information given:
$$22.2740 < \log(2^{74}) < 22.2814$$
$$17.6527 < \log(3^{37}) < 17.6564$$
Hence:
$$39.9267 < \log(12^{37}) < 39.9378$$
Now take $\delta = 40 - 39.9267 = 0.0733$ so that:
$$40 < \log(12^{37}) + \delta < 40.0111$$
Then the first digit of $10^{\log(12^{37}) + \delta}$ must be a 1 (and it must be just above a power of 10) and also:
$$10^{\delta} \approx 1.18$$
Note that:
$$10^{n} / 1 = 10^{n}$$
$$10^{n} / 1.111\cdots = 9 \cdot 10^{n - 1}$$
$$10^{n} / 1.25 = 8 \cdot 10^{n - 1}$$
And 1.18 falls comfortably between 1.11111... and 1.25, so [handwavy error analysis goes here] the first digit of:
$$\frac{10^{\log(12^37) + \delta}}{10^{\delta}} = 10^{\log(12^{37})} = 12^{37}$$
must be an 8.

Bacterius has found

$39.9267 < \log(12^{37}) < 39.9378$
now log 8 < 3 * .3011 or .9033

and log 9 > 2 * .4771 or .9542
as .9378 is between the 2 and also .9267 so the digits is 8.
 
Last edited:
kaliprasad said:
Bacterius has found

$39.9267 < \log(12^{37}) < 39.9378$
now log 8 < 3 * .3011 or .9093

and log 9 > 2 * .4771 or .9542
as .9378 is between the 2 and also .9267 so the digits is 8.

That's much nicer than the second part of my solution! Very nice observation
 
Thanks both for participating in this challenge.

Solution of other:

Note that the given bounds for both $\log 2$ and $\log 3$ allow us to create the following inequalities:

$39+3\log 2<37(2\log 2+\log 3)<39+2\log 3$ and

$8\times 10^{39}<12^{37}<9\times 10^{39}$

Therefore the leftmost digit of the figure $12^{37}$ is $8$.

given $0.3010<\log 2<0.3011$ and $0.4771<\log 3<0.4772$.
 
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