Question

How do you solve `log_12 (x + 1) – log_12 4 = log_12 16`?

Answer 1

Here you need to remember some rules of logarithms!
Consider that:
`loga-logb=log(a/b)`
So you can write:
`log_12((x+1)/4)=log_12(16)`
Also remember that `a^(log_a(x))=x`
So use `12` to eliminat the logs as:
`12^(log_12((x+1)/4))=12^(log_12(16))`
which gives you:
`(x+1)/4=16`
And:
`x=63`