Question

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

Answer 1

I would go for `x=2.8` but this is only an approximation of the real value and also I suspect the base of the second log should be `5` as well....good exercise though!

Explanation:

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....