Why is a logarithmic function one to one? (and exponential function).

Question

$f(x)=\log_a{x}$. This is a logarithmic function. I don't know but I think that it's easier to understand this function if you put it this way, $a^{f(x)}=x$. So one of my questions is that if $f(x_1) \neq f(x_2)$ then $x_1 \neq x_2$, which probably is the same as asking why is the exponential function one to one, and I also wanted to ask why it is an increasing function? And my other question is that if $0<c<1$ (for example) and $b \ge 1$, why is $b^c$ more than one?

Answer 1

First of all they are not one to one on the complex numbers. But that's a different issue. They are as far as real analysis is concerned.

Now if $b>1$ then $b*b>b*1=b>1>\frac 1b>...$. So inductively $b^n $ is increase on the integers.

(If $0 <b <1$ the $b^n $ is decreasing but all arguments and results are equivalent.)

We can extend that to to rationals by noting that $\frac am <\frac cd$ (wolog for positive values) means $ad <mc $ and so $b^{am}=\sqrt [m]{b^a}=\sqrt[mc]{b^{ac}}< \sqrt[ad]{b^{ac}}=\sqrt[d]{b^c}=b^{\frac cd} $.

And for the reals... that depends on how you defined $b^x $ but it is clearly increasing either by limits or by the nature of integrals of positive values (integrals of positive values obviously increase over larger range.)

So $b^x $ increasing.

And increasing function must be 1-1, mustn't they? Think about it. If $x\ne y $ then either $x <y$ or $y <x $. That means $f (x)<f (y)$ or $f (y)<f (x) $.

So $b^x $ is 1-1.

So $\log_b x $ is 1-1 as $b^k=x\iff \log_b x=k $. So if $\log_b x=\log_b y=k $ then $b^k=y $ and $b^k=x $. So $x=y$.