Calculus
g−1(x)=13lnx+53g^{-1}\(\left\)(x\(\right\))=\(\frac\)13\(\ln\) x+\(\frac\)53
g−1(x)=3lnx+5g^{-1}\(\left\)(x\(\right\))=3\(\ln\) x+5
g−1(x)=15lnx+35g^{-1}\(\left\)(x\(\right\))=\(\frac\)15\(\ln\) x+\(\frac\)35
g−1(x)=5lnx+3g^{-1}\(\left\)(x\(\right\))=5\(\ln\) x+3