tolong bantu cara penyelesaian fungsi invers komposisi, susah: makasih

[tex]\displaystyle (g^{-1}\circ f^{-1})(x)=(f\circ g)^{-1}(x)=\frac{2x-5}{5x-1} \\ f(x)=x-2 \\ (f\circ g)(x)=\frac{x-5}{5x-2} \\ g(x)-2=\frac{x-5}{5x-2} \\ g(x)=\frac{x-5}{5x-2}+2=\frac{x-5+10x-4}{5x-2}=\frac{11x-9}{5x-2}[/tex]

No. 6 Skip aja mencurigakan kayaknya gak ada jawabannya,

[tex](g\circ f)(x)=\frac{2f(x)-1}{f(x)} \\ g(x)=\frac{2x-1}{x} \\ f(x-1)=g(x) \\ f(x-1)=\frac{2x-1}{x} \\ f(x)=\frac{2(x+1)-1}{x+1}=\frac{2x+1}{x+1} \\ f^{-1}(x)=\frac{-x+1}{x-2} \\ f^{-1}(3)=\frac{-3+1}{3-2}=-2[/tex]

[tex]f(x)=2^{3x+1},g(x)=^2\log{x}+3 \\ (f^{-1}\circ g^{-1})(x)=(g\circ f)^{-1}(x)=2 \\ (g\circ f)(x)=^2\log{2^{3x+1}}+3=3x+1+3=3x+4 \\ (g\circ f)^{-1}(x)=\frac{x-4}{3}=2 \\ x-4=6 \\ x=10[/tex]

Ati-ati kalo mau minta jawaban banyak-banyak, suka di-delete