What is the value of n in x^n=y if x and y are given?

[univeruser]

Hi all. As the question suggests, there is nothing much I need to explain. It's simple: what is xⁿ=y, if x and n are given? Since this is an algebraic equation, it becomes x=ⁿ√y. Like for example, 3ⁿ=9. With a little bit of common sense, we can see that n is 2, but what about larger numbers?

Please try to keep your method of solving this equation simple, given my inability to understand complicated stuff like logs, antilogs, etc.

If anyone wants to know why I want this, it's because I've been cracking my head on it for weeks. Just like that.

P.S. Please don't use logs. Because, if you look at it carefully, you shall see that this equation itself is a logarithm. x is the base, n the unknown log of y, and y the number.

Better still, tell me how logs can be calculated easily on hand. The same thing, of course.

 

[sylas]

Hi all. As the question suggests, there is nothing much I need to explain. It's simple: what is xⁿ=y, if x and n are given? Since this is an algebraic equation, it becomes x=ⁿ√y. Like for example, 3ⁿ=9. With a little bit of common sense, we can see that n is 2, but what about larger numbers?

I don't understand what you are asking. So if x and n is given, are you asking how to calculate xⁿ?

Or are you asking how to calculate n, given x and y, which is what your numeric example suggests.

In the latter case, you are asking how to calculate the value of the logarithm. That's all logarithm means. A logarithm IS the number to which you must raise a number x (called the base) in order to obtain a given number y.

If you have y = xⁿ, then the equation is rewritten to give n, or x, as follows:

\begin{align*}<br /> y &amp;= x^n \\<br /> x &amp; = y^{1/n} \\<br /> n &amp; = \log_x \; y<br /> \end{align*}​

There's no easy direct way to solve any of these by hand, except in some special cases, such as calculating y or x when n is 2. There are ways to obtain numeric solutions by successive approximations, however. Is that what you want?

Cheers -- sylas

 

[Gib Z]

https://www.physicsforums.com/showpost.php?p=2029082&postcount=4

Thats probably as easy as it'll get.

 

[Redbelly98]

P.S. Please don't use logs.

With this one statement, your request becomes impossible. Or at best impractical, since the only other option would be trial and error.

It's like asking how to solve Ax = B, where A and B are given ... but please don't use multiplication or division!

 

[Mentallic]

But don't you see, it's the lack of simplicity for calculating logs by hand that has forced univeruser to make such a remark.

The difference in your example is that the process of division has rules that make the problem simple to solve by hand. Logs on the other hand are problematic to say the least.

 

[Mark44]

Hi all. As the question suggests, there is nothing much I need to explain. It's simple: what is xⁿ=y, if x and n are given? Since this is an algebraic equation, it becomes x=ⁿ√y. Like for example, 3ⁿ=9. With a little bit of common sense, we can see that n is 2, but what about larger numbers?

The thread title and the first paragraph aren't consistent, and might have lead to some confusion. The title asks what's the value of n if x and y are given, and in the first paragraph, you're asking what is xⁿ = y if x and n are given.

Please try to keep your method of solving this equation simple, given my inability to understand complicated stuff like logs, antilogs, etc.

If anyone wants to know why I want this, it's because I've been cracking my head on it for weeks. Just like that.

P.S. Please don't use logs. Because, if you look at it carefully, you shall see that this equation itself is a logarithm. x is the base, n the unknown log of y, and y the number.

No, an equation can't be a logarithm. The equation is an exponential equation, since the variable you wish to solve for is in the exponent. You're part right, though -- the logarithm of a number is an exponent. More precisely, the logarithm, relative to some base, of a number a is the exponent on the base that produces that number. In mathematical terms. y = log_b a <==> a = b^y.

Better still, tell me how logs can be calculated easily on hand. The same thing, of course.

Generally speaking, logs can't be calculated easily by hand. The same is true for other functions, such as the sine, cosine, and tangent functions. We have ways of calculating values involving these trig functions as well as the logs of numbers in a few bases, and most of them involve power series of one kind or another. The calculations can be done by hand, but they're not what I would call easy.

 

[DrGreg]

Estimating logarithms

As others have already pointed out, if you want an accurate answer for arbitrary values of x, y and n, you have no choice but to use logs, and you'll need a calculator (or, for old-schoolers, book of log tables or slide rule) or else be a masochist for working out a long and complicated power series formula.

However, if you only need an approximate answer, there are ways to make an estimate. And the way you phrased your original question suggests that you might only be interested in the case where n is a positive integer. In this case you can get an approximate answer and round to the nearest integer and with luck that will be the correct answer. (Even if not, you can add or subtract one until you find the right answer, which shouldn't take long.)

So how to get an approximate answer?

Note first that 2¹⁰ = 1024 is quite close to 10³ = 1000, so we can say, approximately,

10³ ≈ 2¹⁰​

so, taking 10th roots,

10^0.3 ≈ 2​

and then, doubling

10^0.6 = 10^2×0.3 ≈ 2² = 4
10^0.9 = 10^3×0.3 ≈ 2³ = 8​

Continuing doubling we get this:

10^0.0 = 1
10^0.3 ≈ 2
10^0.6 ≈ 4
10^0.9 ≈ 8
10^1.2 ≈ 16
10^1.5 ≈ 32
10^1.8 ≈ 64
10^2.1 ≈ 125
10^2.4 ≈ 250
10^2.7 ≈ 500
10^3.0 = 1000
​

Note also that

10^0.1 = 10^2.1 / 10² ≈ 125/100 = 1.25
10^0.2 = 10^1.2 / 10¹ ≈ 16/10 = 1.6
​

and so on, so we get

10^0.0 = 1
10^0.1 ≈ 1.25
10^0.2 ≈ 1.6
10^0.3 ≈ 2
10^0.4 ≈ 2.5
10^0.5 ≈ 3.2
10^0.6 ≈ 4
10^0.7 ≈ 5
10^0.8 ≈ 6.4
10^0.9 ≈ 8
10^1.0 = 10
​

This is enough to estimate any power of ten where the exponent is given to one decimal place, e.g.

10^4.2 = 10^0.2 × 10⁴ ≈ 16000
​

So now let's apply this to an actual problem. E.g. solve 3ⁿ = 2187.

From the above table we can estimate that 3 ≈ 10^0.5 and 2187 ≈ 10^3.3.

So

10^0.5n ≈ 10^3.3
​

and so

0.5n ≈ 3.3
n ≈ 6.6
​

If you were told in advance that n was an integer, you could guess the answer must be 7 and then check that that is indeed correct.

The above technique can be taken further (more accurately) by filling in the table for values in between, e.g. 10^0.05 ≈ 1.1, 10^0.15 ≈ 1.4, ...

 

[Redbelly98]

Thank you DrGreg.

I thought I remembered a similar thread, and after doing a search have found it:

https://www.physicsforums.com/showthread.php?t=283991"​