please, question 4, 5, 7 and 8 urgently, i am in a bit of hurry.thanks

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\log }_{\mathrm{a}}\mathrm{b}=\frac{3}{2} \\ \Rightarrow \mathrm{b}={\mathrm{a}}^{\frac{3}{2}} \\ \Rightarrow \mathrm{a}={\mathrm{b}}^{\frac{2}{3}} \\ {\log }_{\mathrm{c}}\mathrm{d}=\frac{5}{4} \\ \Rightarrow \mathrm{d}={\mathrm{c}}^{\frac{5}{4}} \\ \Rightarrow \mathrm{c}={\mathrm{d}}^{\frac{4}{5}} \\ \mathrm{According} \mathrm{to} \mathrm{question} \\ \mathrm{a}-\mathrm{c}=9 \\ \Rightarrow {\mathrm{b}}^{\frac{2}{3}}-{\mathrm{d}}^{\frac{4}{5}}=9 \\ \Rightarrow {\left({\mathrm{b}}^{\frac{1}{3}}\right)}^2-{\left({\mathrm{d}}^{\frac{2}{5}}\right)}^2=9 \\ \Rightarrow \left[{\mathrm{b}}^{\frac{1}{3}}-{\mathrm{d}}^{\frac{2}{5}}\right]\left[{\mathrm{b}}^{\frac{1}{3}}+{\mathrm{d}}^{\frac{2}{5}}\right]=9 \left[\mathrm{Use} {\mathrm{A}}^2-{\mathrm{B}}^2=\left(\mathrm{A}-\mathrm{B}\right)\left(\mathrm{A}+\mathrm{B}\right)\right] \\ \Rightarrow \left[{\mathrm{b}}^{\frac{1}{3}}-{\mathrm{d}}^{\frac{2}{5}}\right]\left[{\mathrm{b}}^{\frac{1}{3}}+{\mathrm{d}}^{\frac{2}{5}}\right]=1\times 9 \\ \mathrm{Take} \\ {\mathrm{b}}^{\frac{1}{3}}-{\mathrm{d}}^{\frac{2}{5}}=1...\left(\mathrm{i}\right) \\ \mathrm{and} \\ {\mathrm{b}}^{\frac{1}{3}}+{\mathrm{d}}^{\frac{2}{5}}=9...\left(\mathrm{ii}\right) \\ \left(\mathrm{i}\right)+\left(\mathrm{ii}\right) \\ {\mathrm{b}}^{\frac{1}{3}}-{\mathrm{d}}^{\frac{2}{5}}+{\mathrm{b}}^{\frac{1}{3}}+{\mathrm{d}}^{\frac{2}{5}}=1+9 \\ \Rightarrow 2{\mathrm{b}}^{\frac{1}{3}}=10 \\ \Rightarrow {\mathrm{b}}^{\frac{1}{3}}=5 \\ \Rightarrow \mathrm{b}=5^3 \\ \Rightarrow \mathrm{b}=125 \\ \mathrm{Put} {\mathrm{b}}^{\frac{1}{3}}=5 \mathrm{in} \left(\mathrm{i}\right) \\ 5-{\mathrm{d}}^{\frac{2}{5}}=1 \\ \Rightarrow {\mathrm{d}}^{\frac{2}{5}}=4=2^2 \\ \Rightarrow {\mathrm{d}}^{\frac{1}{5}}=2 \\ \Rightarrow \mathrm{d}=2^5 \\ \Rightarrow \mathrm{d}=32 \\ \mathrm{b}-\mathrm{d}=125-32=93 \left[\mathrm{Answer}\right] \\ \mathrm{Please} \mathrm{ask} \mathrm{one} \mathrm{question} \mathrm{in} \mathrm{one} \mathrm{thread}. \end{gathered}$$

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