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

I wanted to provide a graphical approach to this problem...

The answer to this problem is where the function, #y = log_5 (x+1) - log_4 (x-2) # interects with #y = 1 #:

Hence from this graph we see that #x approx 2.787 # is a solution for this...

Now i understand that this is not the most ellegant solution, and not the traditional method, so i want to hopefully start a more rigerouse approach:

We could possibly use a very forgotten identity in logarithms...

#log_alpha beta -= (log_phi beta) / (log_phi alpha) #

If you have never seen this before, its a good idea to add this to your inventory of identities...

Applying this...

#=> log(x+1)/log5 - log(x-2)/log4 = 1 #

#=> log4 *log(x+1) - log5 *log(x-2) = log4 *log 5 #

#=> log((x+1)^(log4) ) - log((x-2)^(log5)) = log4 * log5 #

#=> log( (x+1)^(log4) / (x-2)^(log5) ) = log4 * log 5 #

#=> (x+1)^(log4) / (x-2)^(log5) = 10^ ( log4 * log 5 ) #

#=> (x+1)^(log4) = 10^( log4 * log5 ) * (x-2)^(log5) #

Now maybe we can use binomial epxansion and other approximations, but i hope this was a good start to one apporach, but maybe there are more simple approaches, hopefully someone else can provide such a method!