If you mean "log<sub>b</sub>(s/t)", then, yes, the "expanded" form would be "log<sub>b</sub>(s) - log<sub>b</sub>(t)".Elaine said:express logbs/t as a difference of logarithms Is this solution logbt-logbs?
There are no exponents in your answer, which should tip you off that there may be a problem here.Elaine said:convert logn T=-x to an expoenetial equation Is this n-x=T???
You have it backwards.Express log<sub>b</sub>(s/t) as a difference of logarithms
Is this solution: log<sub>b</sub>(t) - log<sub>b</sub>(s) ? . . . . no
If that is what you meant to write, it's correct.Convert log<sub>n</sub>(T) = -x to an expoenetial equation
Is this: n<sup>-x</sup> = T ?