In the finite field $\mathbb{F}_{101}$ ,where discrete logarithms are $L_2(3)=69$ and $L_2(5)=24$. Compute the discrete logarithm $ L_2(60)$?

Question

Now, I have that $L_2(60)=L_2(4*3*5)=L_2(4)+L_2(3)+L_2(5)=2+69+24=95$. So from my work $L_2(60)$ is $95$, but the answer on some other website gives $14$. I just don't see where I went wrong.

Answer 1

Write $\equiv$ for "equal modulo $101$". You are told that $2^{69}\equiv 3$ and $2^{24}\equiv5$. Furthermore $2^2\equiv4$ trivially. From $60=3\cdot 5\cdot 4$ it then follows that $$60\equiv2^{69}\cdot2^{24}\cdot2^2\equiv2^{95}\ .$$ This shows that in ${\mathbb F}_{101}$ one has $L_2(60)=95$, whereby we have not checked that $2$ is actually a generator of the multiplicative group ${\mathbb F}_{101}^*$, vulgo: a primitive root.