Question

Question #63922

Answer 1

I found `x=15` assuming logs in base `10`.

Explanation:

Let us rearrange it and write:
`log(2)-log(x+5)=-1`
We use a property of logs to change the dfference of logs into a log of a fraction:
`log(2/(x+5))=-1`
Now the problem is the base of the log; I am going to assume it is `10`*:
`log_(10)(2/(x+5))=-1`
Using the definition of log we write:
`2/(x+5)=10^(-1)`
Rearrange:
`x+5=20`
`x=15`

*If the base is not `10` you can use the same method substituting into the last steps.