Question

Let logba=3 logbc=2 logbd=5 what is the value of logba^3d^4/c^2

Answer 1

25

Explanation:

Answer 2

`Use:\\\log_ax+\log_ay=\log_a(xy)\\\\\log_ax-\log_ay=\log_a(x)/(y)\\\\\log_ax^n=n\cdot\log_ax`


`\log_ba=3;\ \log_bc=2;\ \log_bd=5`


`\log_b(a^3d^4)/(c^2)=\log_b(a^3d^4)-\log_bc^2=\log_ba^3+\log_bd^4-\log_bc^2\\\\=3\log_ba+4\log_bd-2\log_bc=3\cdot3+4\cdot5-2\cdot2=9+20-4=25`