First, use the property that #log_{a}(C/D)=log_{a}(C)-log_{a}(D)# to rewrite the equation as
#log_{a}((X+4)/(X-7))=log_{a}((X-8)/(X-9))#
Next, use the fact that #log_{a}(z)# is a one-to-one function (when #a>0# and #a!=1#) to say that the inputs are the same:
#(X+4)/(X-7)=(X-8)/(X-9)#
Now cross-multiply to get:
#(X+4)(X-9)=(X-8)(X-7)#
After using FOIL, this becomes
#X^{2}-5X-36=X^2-15X+56#.
Rearranging and canceling appropriately gives
#10X=92# so that #X=92/10=46/5=9.2#
You should always check your answer in the original equation. When #X=9.2#, we have
#X+4=13.2#, #X-7=2.2#, #X-8=1.2#, and #X-9=0.2#
You can use any value for #a>0# except #a=1#. For instance, if we use #a=10#, then:
#log_{10}(13.2)approx 1.120574#, #log_{10}(2.2)approx 0.342423#, #log_{10}(1.2)approx 0.079181#, and #log_{10}(0.2)approx -0.698970#
and #log_{10}(13.2)-log_{10}(2.2)approx 1.120574-0.342423= 0.778151# and #log_{10}(1.2)-log_{10}(0.2)approx 0.079181-(-0.698970)=0.778151#
