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

[anemone]

Find the leftmost digit of the figure $12^{37}$ given $0.3010<\log 2<0.3011$ and $0.4771<\log 3<0.4772$.

 

[Nono713]

Spoiler

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.

 

[kaliprasad]

Spoiler

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

Spoiler

$39.9267 < \log(12^{37}) < 39.9378$
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.

 

[Nono713]

Bacterius has found

Spoiler

$39.9267 < \log(12^{37}) < 39.9378$
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

 

[anemone]

Thanks both for participating in this challenge.

Solution of other:

Spoiler

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 $0.4771<\log 3<0.4772$.