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I am curious to know the concept behind. Let's say y = 10^x So, y should be a number with zeroes but y is 10 multiplied to 10 x times. But this is not the case.

For eg 10^3.5 = 3162.27766017

So, not a number ending with zeroes. I know the mathematics behind this but I would like to know the concept and reasoning as to how I might explain to a high school student why does 10 to the power something ends up with a number that may not end with zeroes ?

This might be a silly question but I would appreciate if I can get some answer before this question is put on hold/etc.

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marked as duplicate by JMoravitz, Xander Henderson, Leucippus, Shailesh, José Carlos Santos Oct 10 at 9:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $y$ will have zeroes if $x$ is a natural number, not in general if $x$ is a real number. Consider $\sqrt{10} = 10^{1/2}$ $\endgroup$ – Jack Pfaffinger Oct 9 at 18:28
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    $\begingroup$ What does multiplying $10$ by itself three and half times mean? Why should that end with zeroes? $\endgroup$ – saulspatz Oct 9 at 18:32
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You can write $$\log_{a}{x}=\frac{\ln(x)}{\ln(a)}$$ where $$a>0,a\neq 1$$ P.s.: $\ln(x)$ Is the Logarithmus naturalis.

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  • $\begingroup$ Oh. I always thought it was log Napierian. $\endgroup$ – Lubin Oct 9 at 18:37
  • $\begingroup$ One can also write $$\log_{a}{x}=\frac{\lg(x)}{\lg(a)}$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 9 at 18:38
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Take any number say $567$. Suppose we to find a number $x$ such that $10^x = 567$. How will we do it?

We know that $10^2 < 567 < 10^3$ hence $x$ must lie between $2$ and $3$. Also $10^x$ is a strictly increasing function and continuous of $x$ and it will take all values from $0$ to $\infty$ if we vary $x$ from $-\infty$ to $\infty$. Hence there must be exactly one value of $x$ between $2$ and $3$ for which $10^x = 567$. We call this unique value of $x$ as the logarithm of $567$ to the base 10.

Now it may so happen that this particular value of $x$ is a irrational number with never ending decimals.

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Assume $a>0$.

In general, if $x$ is not an integer, it does not make sense to say $$a^x=\underbrace{a\cdot a\cdots a}_{x\text{ times}},$$ because you can't multiply something by itself a non-integer number of times. Instead, we define the function $f(x)=a^x$ by, for any $x$, $$f(x+1)=a\cdot f(x), \qquad f(0)=1,$$ so that the familiar definition of exponentiation is satisfied for integer $x$.

Thus, we need only define $a^x$ for $0<x<1$, because if $s>1$, there is an integer $n$ such that $s=x+n$ and $0<x<1$. In this case, we would have $$f(s)=f(x+n)=a^nf(x).$$

"Unfortunately", the definition for $0<x<1$ requires a little bit of calculus.

Since you are using logarithms, I'm going to assume you know what the number $e$ is. $e$ has some remarkable properties. One such property is the definition of the exponential function:

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\dots.$$ Since $a=e^{\ln a}$, we see that $a^x=e^{x\ln a}$: $$a^x=1+\frac{x\ln a}{1!}+\frac{(x\ln a)^2}{2!}+\frac{(x\ln a)^3}{3!}+\frac{(x\ln a)^4}{4!}+\frac{(x\ln a)^5}{5!}+\dots.$$ Such a definition will work.

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It's an extension of the usual laws of exponents for positive integers.

If we want some basic properties to be satisfied, then there is a way to extend this to negative integers, and zero. Thus, we can define integer powers of a positive number, like $10.$

Next come fractional indices, which are a bit subtle, but we may still define fractional powers of positive numbers in a way consistent with out extension to integer powers.

We can also extend to arbitrary real indices, but I think for your case it is sufficient to stop at rational indices.

So, you see, we can actually have such things as $10^{0.73578},$ and it is easy to see why such quantities cannot again be whole numbers. Basically the idea is that the function $10^x$ is monotonic. That is, it always increases as $x$ is increased. This is true for any positive base apart from $1,$ so there's nothing special about the $10.$ (The only caveat is that for bases less than $1,$ we have that $b^x$ decreases with increasing $x.$) Now, after having seen this, note that $10^0=1$ and $10^1=10.$ Thus, if you pick any index $i$ between $0$ and $1$ (there are infinitely many of them), you shouldn't expect too much by way of an integral value for $10^i$ since there is a finite number of integers between $1$ and $10.$ In fact, most of such values are irrational. Your question of why some powers of $10$ can fail to be integral is thus answered -- the idea is that that's what falls out of our extension of the idea of positive integral indices to arbitrary exponents, and it follows by the necessary fact that in any interval we always have infinitely many non-integers and finitely many integers. So it's a consequence of how we've chosen to define the function $10^x$ and its siblings, and the pigeonhole principle.

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Note that positive integer powers of $10$ end up with zeros.

$$10^1=10, 10^2=100, 10^3=1000, $$

Thus for example $$10^{2.5} = 316.2277660....$$ is a number between $100$ and $1000$

Speaking of logarithm, if for example you want to find a power $x$ for which $10^x =20$ you take logarithm of $20$

$$\log 20 = 1.301029996...\iff 10^{1.301029996...}=20 $$

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  • $\begingroup$ That's what I am asking ! Why does 10^(something) end up being 20/3162.27766/etc ? $\endgroup$ – Sristy Oct 9 at 18:42
  • $\begingroup$ @Sristy this is the consequence of the intermediate value theorem. Once you know the function $10^x$ is continuous and $10^2=100$ and $ 10^3=1000$, the intermediate value theorem indicates that everything between $100$ and $1000$ will be attained by a number between $2$ and $3$. $\endgroup$ – Mohammad Riazi-Kermani Oct 9 at 19:11
  • $\begingroup$ According to my calculator, $10^{2.5}=310.227...$ $\endgroup$ – J. W. Tanner Oct 9 at 19:57
  • $\begingroup$ @J.W.Tanner Please check it again, it is $316.2277660...$ $\endgroup$ – Mohammad Riazi-Kermani Oct 9 at 20:30
  • $\begingroup$ @MohammadRiazi-Kermani: yes, I meant $316.227...$; I see you corrected $\endgroup$ – J. W. Tanner Oct 9 at 20:33

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