$F(xy) = F(x)+F(y)$ Proof

Question

Suppose $F$ is differentiable $\forall x>0$ and $F(xy) = F(x)+F(y)$, $ \forall x,y>0$. Prove that if $F$ is not the zero function, then $\exists$ $ a>0$ such that: $F(x)=\log_a(x)$, $\forall x>0$. I seem to be doing fine except on getting the base $a$ for the log. So far I have that $F'(x)=\frac{F'(1)}{x}$. I know from calculus that $\int\frac{1}{t}dt=\ln(t)$. How can I get the base to be $F'(1)$ instead of $e$?

Answer 1

Since $F(xy)=F(x)+F(y)\implies F(1)=0$ we have $F(x)=F(x)-F(1)=\int_1^x (F'(1)/t)dt=F'(1)\ln x =\log_A(x)$ where $$A=\exp {(1/F'(1))}.$$ Because for $x>1$ we have $$\exp(\ln x)=x=A^{\log_A(x)}=[\exp (\ln A)]^{\log_A(x)}=$$ $$=\exp (\;(\ln A)(\log_A(x)\;).$$ So $\log_A(x)=(\ln x)\cdot (1/\ln A).$

In this case we have $1/\ln A=F'(1),$ that is, $\ln A=1/F'(1).$

For example if $F'(1)=10$ the base $A$ is $e^{1/10}.$

Answer 2

You can conclude that$$F(x)=F'(1)\ln x+C$$where $C=0$ since $F(1)=0$ ( let $x=y=1$ ) therefore $$F(x)=F'(1)\ln x={\log_e x\over \log_e e^{F'(1)}}=\log_{e^{F'(x)}}x$$hence you can write $$a=e^{F'(1)}$$ where $F'(1)$ is arbitrarily any real number.

Answer 3

$$a^x=\left(e^{\text{log}(a)}\right)^x = e^{\text{log}(a)x}$$ $$\text{log}(a^x)=x\text{log}(a)$$ $$\text{log}_a(a^x)=x$$

Extrapolating from here, we have

$$\text{log}_a(x) = \frac {\text{log}(x)}{\text{log}(a)}$$

It shouldn't be too hard to get your answer from this.