Playing with e and ln

[Jason76]

We know \(\displaystyle e^{0} = 1\), but how to prove it?

Considering function \(\displaystyle y = e^{x}\) where \(\displaystyle y = 1\)

\(\displaystyle 1 = e^{x}\) - Solve for \(\displaystyle x\).

\(\displaystyle \ln(1) = \ln(e^{x})\) - Using \(\displaystyle \ln\) to get an ordinary \(\displaystyle x\) from \(\displaystyle e\).

\(\displaystyle 0 = x\)

Plugging back into the equation:

\(\displaystyle 1 = e^{0}\) or \(\displaystyle e^{0} = 1\)

We know \(\displaystyle \ln(1) = 0\), but how to prove it?

Considering function \(\displaystyle y = \ln(x)\) where \(\displaystyle y = 0\)

\(\displaystyle 0 = \ln(x)\) - Solve for \(\displaystyle x\).

\(\displaystyle e^{0} = e^{\ln(x)}\) - Using \(\displaystyle e\) to get an ordinary \(\displaystyle x\) from \(\displaystyle \ln\).

\(\displaystyle 1 = x\)

Plugging back into the equation:

\(\displaystyle 0 = \ln(1)\) or \(\displaystyle \ln(1) = 0\)

 

[daon2]

\(\displaystyle e^0=1\) can be thought of as a definition to make the function \(\displaystyle f(x)=e^x\) continuous at \(\displaystyle x=0\). It is hard to justify intuitively from the definition of an integral power. It is sometimes referred to as the empty product. Since \(\displaystyle \ln(x)\) is defined to be the inverse function to \(\displaystyle e^x\), we guarantee \(\displaystyle \ln(1)=0\).

You can also justify it by first assuming the exponential rule \(\displaystyle e^{x-y}=e^x/ e^y\) and substituting in \(\displaystyle y=x\).

 

[stapel]

We know \(\displaystyle e^{0} = 1\), but how to prove it?

The natural exponential, when raised to the power zero, is the same value as every other (non-zero) value when raised to the power zero: 1. And for the same reason, with the same "proof".

We know \(\displaystyle \ln(1) = 0\), but how to prove it?

By using the definition of logs: "ln(1)" is the power which, when put on the base (e), will result in a value of 1. Since this power is zero, because e^0 = 1, then the value of ln(1) must be zero.

 

[Jason76]

I wonder how you could show (algebraically by solving for \(\displaystyle x\)) \(\displaystyle 1 = a^{0}\), considering the equation \(\displaystyle y = a^{x}\).

Starting out as: \(\displaystyle 1 = a^{x}\).

 

[pka]

I wonder how you could show (algebraically by solving for \(\displaystyle x\)) \(\displaystyle 1 = a^{0}\), considering the equation \(\displaystyle y = a^{x}\).
Starting out as: \(\displaystyle 1 = a^{x}\).

If \(\displaystyle a\ne 0\), would you agree that \(\displaystyle \dfrac{a}{a}=1~?\).

Would you agree that \(\displaystyle \dfrac{a^x}{a^y}=a^{x-y}~?\)

Then would you agree that \(\displaystyle 1=\dfrac{a^x}{a^x}=a^{x-x}=a^0~?\)

 

[Jason76]

If \(\displaystyle a\ne 0\), would you agree that \(\displaystyle \dfrac{a}{a}=1~?\).

Would you agree that \(\displaystyle \dfrac{a^x}{a^y}=a^{x-y}~?\)

Then would you agree that \(\displaystyle 1=\dfrac{a^x}{a^x}=a^{x-x}=a^0~?\)

Right, I agree. That's another way of looking at it.

 

[pka]

Right, I agree. That's another way of looking at it.

But why?

If \(\displaystyle a\ne 0\) then \(\displaystyle a^x=1\text{ if and only if } x=0\).

The is the basis of what you are asking.

 

[HallsofIvy]

How, exactly, are you defining \(\displaystyle a^x\)?

One way it to start with x a positive integer: then \(\displaystyle a^x\) is defined as "a multiplied by itself x times". We can show, from that definition that \(\displaystyle a^{x+ y}= a^xa^y\) and \(\displaystyle a^{xy}= (a^x)^y\) for x and y any positive integers. We then extend that definition to other numbers, all integer, rational numbers, real numbers, so that those properties remain true. For example, if we want \(\displaystyle a^{x+ y}= a^xa^y\) to be true even when y= 0, we must have \(\displaystyle a^{x+ 0}= a^xa^0\) which, since x+ 0= x, is the same as \(\displaystyle a^x= a^xa^0\). Since \(\displaystyle a^x\ne 0\) for any positive integer x, we can divide both sides by \(\displaystyle a^x\) to get \(\displaystyle 1= a^0\). That is, in order that \(\displaystyle a^{x+y}= a^xa^y\) even when y= 0, we must define \(\displaystyle a^0= 1\).

Another way to define \(\displaystyle a^x\) is to start by defining the natural logarithm as an integral:
\(\displaystyle ln(x)= \int_1^x \frac{1}{t}dt\)
and then define \(\displaystyle e^x\) as its inverse function. Since \(\displaystyle ln(1)= \int_1^1\frac{1}{t}dt= 0\), we must have \(\displaystyle e^0= 1\). And then, since \(\displaystyle a^x= e^{ln(a^x)}= e^{x ln(a)}\), it follows that \(\displaystyle a^0= e^{0 ln(a)}= e^0= 1\).

 

[lookagain]

But why?

If \(\displaystyle a\ne 0\) then \(\displaystyle a^x=1\text{ if and only if } x=0 \ \ \ \). That is a double implication.

The is the basis of what you are asking.

Yes, if \(\displaystyle \ a \ne 0 \ \ and \ \ x = 0, \ \ then \ \ a^x = 1. \ \ \) But, \(\displaystyle \ if \ \ a \ne 0 \ \ and \ \ a^x = 1, \ \ then \ \ a = 1 \ \ and \ \ x = \ \ any \ \ real \ \ number, \ \ or \ \ a = -1 \ \ and \ \ x = \ any \ \ even \ \ integer.\)