Question

Solve for `x`: `log(x) - log(x + 7) = -1`?

Answer 1

`x=7/9`.

Explanation:

Remember the rules of logarithms: `loga-logb=log(a/b)`. Using this, we get

`logx-log(x+7)=-1`
`=>log(x/(x+7))=-1`

Now we remember that `logc=d<=>c=10^d`. Essentially what we're doing there is taking 10 to the power of both sides, and since `10^x` and `logx` are inverse functions, they cancel off. `f^-1[f(x)]=x`. So we get

`log(x/(x+7))=-1`
`(=>10^(log(x/(x+7)))=10^-1)`
`=>x/(x+7)=10^-1`
`=>x=(x+7)/10`
`=>10x=x+7`
`=>9x=7`

`=>x=7/9`

So `x=7//9` is our answer.

Answer 2

Please see the explanation for steps leading to the answer: `x = 7/9`

Explanation:

Given: `log(x) - log(x + 7) = -1`

Use the property `log(a) - log(b) = log(a/b)`

`log(x/(x + 7)) = -1`

Make logarithm disappear by making both sides the exponent of 10:

`cancel10^(cancel(log)(x/(x + 7))) = 10^(-1)`

Making a logarithm an exponent of its base, makes both disappear:

`x/(x + 7) = 10^(-1)`

Multiply both sides by `10(x + 7)`:

`10cancel((x + 7))x/cancel(x + 7) = cancel10(x + 7)cancel10^-1`

`10x = x + 7`

`9x = 7`

`x = 7/9`

Check:

`log(7/9) - log(7/9 + 7) = -1`

`-1 = -1`
This checks.