Question

How do you solve `log_2(x)+log_3(x+1) = 2`?

Answer 1

`=>x = 2`

Explanation:

Rearrange;

`Log_3 (x+1) = 2 - log_2 x`

`Log_2 (x+1) = log_2 4 - log_2 x`

`Log_3 (x+1) = log_2 (4/x)`

Now lets convert everything to base 10

`log(x+1)/log3 = log(4/x)/log2`

`log (x+1)log 2 = log(4/x)log 3`

Now I am going to take all the x terms and dump them on one side and the numerical terms on one side

`log(x+1)/log(4/x) = log3/log2`

Now we can use change of base formula
`log_(4/x) (x+1) = log_2 3`

`=>4/x = 2`

`=>x = 2`

`=>x + 1 = 3`

`=>x = 2`

Technically this step isn't valid for all logs as the bases need not be the same to have the same log value
But as you can see the basis and whats on top yield the same value of x hence this is the solution
We still dont know 2 is the only solution
Yes it is , you need limiting and differential calculus which is wasteful at this point in time

Hence we arrive at a unique solution of 2