why can we use log instead of ln

for (g) I got lg( 1/y-x^2)= lg (a^b-x) how do I continue

Greetings. The choice of employing log or ln is purely arbitrary. Both can be used. If a natural exponential term appears in the equation, ln is of course preferred.

For

lg(1/y - x^2) = (b - x) lg a

lg(1/y - x^2) = - lg a . x + b lg a

so both method are the same?

Affirmative. The graph drawn using lg will however be different from that drawn with ln in terms of the numeric values of the gradients and vertical axis intercepts. The values of the constants thus derived from the respective graphs will still be the same.

okay thank you very much

My pleasure. Regards