Question

How do you solve `log_b30-log_b5^2=log_bx`?

Answer 1

`x=6/5`

Explanation:

`log_b 30-log_b 5^2=log_b x ?`

`"so ;" log _a( b/c)=log_a b-log_a c`

`log _b (30/25)=log _b x`

`30/25=x`

`x=6/5`