How do you solve #log_a (X+4) - log_a (X-7) = log_a (X-8) - log_a (X-9)#?

2 Answers
Jul 24, 2015

#X=46/5=9.2#

Explanation:

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#

Jul 24, 2015

I found #x=9.2#

Explanation:

You can use the fact that: #log_ax-log_ay=log_a(x/y)# to get:
#log_a((x+4)/(x-7))=log_a((x-8)/(x-9))#
take the power of #a# on bothe sides to get rid of the logs:
#cancel(a)^(cancel(log_a)((x+4)/(x-7)))=cancel(a)^(cancel(log_a)((x-8)/(x-9)))#
#(x+4)/(x-7)=(x-8)/(x-9)#
#(x+4)(x-9)=(x-8)(x-7)#
#cancel(x^2)-9x+4x-36=cancel(x^2)-7x-8x+56#
#10x=92#
#x=9.2#