How to find a base and exponent of a specific given number?

Question

I came across this specific problem today in the class that what could be the base and exponent of $8192$ I know that it can be guessed that the base is $2$ and exponent is $13$ but I need a proper solution using algorithm.

Your answers are appreciated!

Answer 1

Presumably you are insisting that the base and exponent be naturals. Otherwise we could also write $8192\approx \pi^{7.87165}$ or many other choices.

You should start by finding some factor of your number. For $8192$ it is obvious $2$ is a factor. You can then compute $\frac {\log(8192)}{\log(2)}=13$ and you are done. If your number is not a power of that factor, the division will not come out even. Say you were given $7776=6^5$. Again you might notice $2$ is a factor but then would find $\frac {\log(7776)}{\log(2)}\approx 12.925$ You can then divide $2$ into $7776$ as many times as possible, here getting $7776=2^5\cdot 243$ The next try is to assume the number is a fifth power, so you can do $243^{1/5}=3$ and you are done with $7776=(2\cdot 3)^5$ Finally you might find that the root doesn't come out well. Say you are given $9834496$. You would find $2$ as a factor, and $9834496=2^{12}\cdot 2401, 2401^{1/12}\approx 1.913$. The correct exponent must be a factor of $12$, so just try them until you find $7^4=2401, 9834496=(2^3\cdot 7)^4$. Note that the answer may not be unique because you could also write $9834496=(2^6\cdot7^2)^2$

If you can't find a small factor easily, the power can't be very large, so another approach is just to take roots until you find the answer. For example $26439622160671=31^9$. If you just start taking roots you don't have to look far. In this case the cube root would come out even because $26439622160671=29791^3$ If you want the highest power you could restrict yourself to multiples of $3$ because you know the number is a cube.

Answer 2

You don't have to guess. Since $8\,192$ is even, you can divide it by $2$, getting $8\,192=2\times4\,096$. Then you do the same thing with $4\,096$, getting $4\,096=2\times2\,048$. And so on… After $13$ divisions by $2$ you get $1$ and therefore $8\,192=2^{13}$.

Answer 3

So if I understand what you mean by “find the base and exponent of a specific number,” the problem is,

Given a positive integer $c$, find positive integers $a$ and $b$ such that $c=a^b$.

Factoring into primes is the key. For your example $8192 = 2^{13}$. So $a=2$ and $b=13$ is a solution. Are there others? Any “base” $a$ must be a factor of $8192$, which (because the prime factorization is unique), must be a power of $2$. So $a=2^k$ for some $k$. But then $a^b = (2^k)^b = 2^{bk}$, which, again by uniqueness, means that $k$ is a factor of $13$. But $13$ is prime so $k=1$ or $k=13$. $k=1$ is the case we've already discovered ($a=2$ and $b=13$). $k=13$ is a sort of trivial case $a=8192$ and $b=1$; we probably want to exclude that.

But we may not always get a unique nontrivial $a$ and $b$. For instance, try $c=64$. We can factor it into $2^6$. The factors of $6$ besides $1$ and $6$ are $2$ and $3$. Accordingly, $64 = (2^3)^2=8^2$ and $(2^2)^3=4^3$. So there are three possibilities:

• $a=2$ and $b=6$
• $a=4$ and $b=3$
• $a=8$ and $b=2$.

Answer 4

For $8192$, José told you how to do it, and this method is just actually writing the unique prime factorization of this given number (note that not every number can be written in the form $a^b$ where $a$ and $b$ are natural numbers).

For example $12=2^2×3$ (this is its unique prime factorization) and to get it you start thinking the same way José told you, first divide $12$ by $2$ to get $6$, and again by $2$ to get $3$ and finally by $3$ to get $1$. (also note that $12 \ne a^b$).

In some cases you might be able to write a number as $a^b$ like for example $36=2^2×3^2=6^2$ (since the exponent is the same for both bases).

Answer 5

8192/2=4096

4096/2=2048

2048/2=1024

1024/2=512

512/2=256

256/2=128

128/2=64

64/2=32

32/2=16

16/2=8

8/2=4

4/2=2

2/2=1

2^13=8192

we are dividing it by 2 so base is 2 We divided it 13 times to get 1 so exponent is 13

Answer 6

The base could be $10$ and exponent could be $\log 8192$.

Or the base could be $e$ and the exponent could be $\ln 8192$.

Or the base could be any $b >0; b\ne 1$ and the exponent $\ln b\cdot \ln 8192$.

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Or the exponent could be $2$ and the base could be $\sqrt {8192}$.

Or the exponent could be $5$ and the base could be $\sqrt[5]{8192}$.

Or the exponent could be any integer $k \ge 1$ the base could be $\sqrt[k]{8192}$.

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I have to wonder if this wasn't what the point of the question was.

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But if it is assumed that the base and exponents both have to be integers then......

Every number has a unique prime factorization.

If $N = p_1^{a_2}p_2^{a_2}.....p_n^{a_n}$ where $p_i$ are prime numbers and $a_i$ are integer powers....

then if $N = b^k$ then $b = p_1^{\frac {a_2} k}p_2^{\frac {a_2} k}.....p_n^{\frac {a_n} k}$ and that can only happen if $k$ is a common divisor of all $a_i$ powers.

So if $N= 8192$ we must find the prime factorization of $8192$. And by repeatedly dividing by $2$ we find $8192 = 2^{13}$.

So if $N = b^k$ and $b$ and $k$ are both integers, then $b = (2^{\frac {13}k})$ where $k$ divides $13$.

But $13$ is prime so $k = 1$ or $k=13$ and that gives us two solutions.

$k = 1$ and $b = 2^{13} = 8192$ and $8192 = 8192^1$ or $k=13$ and $b=2$ and $8192 = 2^{13}$.

....

Let's try another let $N = 80621568$.

First find the prime factorization of $80621568$. By successively dividing by $2$ we find $80621568= 2^{12}* 19683$. And by successively dividing by $3$ we find $19683 = 3^9$ and so $806281568 = 2^{12}*3^9$.

So if $806281568 = b^k$ we must have $b = 2^{\frac {12}k}*3^{\frac {9}k}$ and $k$ must divide both $12$ and $9$. The common factors of $12$ and $9$ are $1$ and $3$ so we have two options:

$k = 1$ and $b=2^{12}3^9=806281568$ and $806281568 =806281568^1$ or $k =3$ or

$k =3 $ and $b=2^{\frac {12}3}3^{\frac 93} = 2^4*3^3 = 16*27=432$ and $806281568 = 432^3$.