Let's revisit the theory of logarithms before we address this problem.

#log_a(x)# (where a>0 and x>0) is a number that, if used as an exponent with a base #a#, produces #x#.

That is, **by definition of the logarithm**,

#a^(log_a(x)) = x#

As is well known from the properties of the *exponential function*, #a^(p+q)=a^p*a^q#

Using this for #p=log_a(x)# and #q=log_a(y)#, we get the following:

#a^[log_a(x)+log_a(y)] = a^[log_a(x)] * a^[log_a(y)] = x*y = a^[log_a(x*y)]#

Therefore,

#log_a(x)+log_a(y) = log_a(x*y)#

As a consequence from this,

#log_a(x)+log_a(1/x) = log_a[x*(1/x)] = log_a(1) = 0#

Therefore,

#log_a(1/x) = -log_a(x)#

From this we also can derive the following:

#log_a(x/y) = log_a[x*(1/y)] = log_a(x)+log_a(1/y) = #

#= log_a(x)-log_a(y)#

Using all this theory, we can calculate

#log_8(7)-log_8(s)+log_8(t)-log_8(4) = #

#= log_8(7/s)+log_8(t/4) = log_8((7t)/(4s))#