How do you solve #log_5(x+1) - log_4(x-2) = 1 #?

We can try changing base to the second log and write it as:
#(log_5(x-2))/(log_5(4))#
so the equation becomes:
#log_5(x+1)-(log_5(x-2))/(log_5(4))=1#
to simplify our evaluation let us write #log_5(4)=k#:
#log_5(x+1)-(log_5(x-2))/k=1#
rearranging:
#klog_5(x+1)-(log_5(x-2))=k#
using a first property of logs:
#log_5(x+1)^k-(log_5(x-2))=k#
a second ne:
#log_5[(x+1)^k/(x-2)]=k#
and the defintion of log:
#(x+1)^k/(x-2)=5^k#
remembering that #log_5(4)=k# we have:
#(x+1)^k/(x-2)=4#
and:
#(x+1)^k=4x-8#
but...the problem is #k#....I got that:
#log_5(4)=k=0.86135#
to keep on going I will try to approximate and write that #k=1#...ok it is not right but let us try it.
we get:
#(x+1)^1=4x-8#
rearranging:
#x+1=4x-8#
#3x=9#
#x=9/3=3+E#
where #E# is the error introduced by my approximation.

Let us test our result into the original equation:
#log_5(4)-log_4(1)=1#
#0.86135-0=1# more or less....

I tried various values for the Error #E# and I found that if #E=-0.2#
we get that:
#x=3-0.2=2.8#
giving in the test of our equation with #x=2.8#:
#log_5(2.8+1)-log_4(2.8-2)=1#
#0.99=1# I think it is ok....