Question

How do you solve `logx+log5=1`?

Answer 1

`x = 2` (Assuming `Log` is `Log_ 10`)

Explanation:

`Log(x) + Log(5) = 1`

Applying `Log(a) + Log(b) = Log(ab)`
`-> Log(5x) = 1`
Assuming `Log` is `Log_10`
`-> 5x = 10^1`
`5x = 10`
`x = 2`