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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 $intfrac{1}{t}dt=ln(t)$. How can I get the base to be $F'(1)$ instead of $e$?

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 $e^{1/10}.$

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

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.

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