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))#
