Question

How do you solve `log(x-3)+log x=1`?

Answer 1

5

Explanation:

`log(x−3)+logx=1`

logarithms on the left side may be added together by multiplying argumnts, on the other side we can rewrite number 1

`log[(x−3)*x]=log10`

Logarithm is a simple function, therefore we can compare arguments

`[(x−3)*x]=10`

`x^2−3x-10=0`

`(x+2)*(x-5)=0`

`x_1=-2`
`x_2=5`

Since argument of logaritm can be only positive, `x_1` is not a solution
`log(-2−3)+log(-2)=>` not possible

`log(5−3)+log5=>log2+log5=>log(5*2)=>log10=>1`
`1=1` ;correct;

Answer 2

See below

Explanation:

The goal with type of problems is to get a expresion like `logA=logB`. By injectivity of function log, we can say that `A=B`

Let see...
Using logarithmic rules

`log(x-3)+logx=log10=1`

`log(x-3)x=log10`

Then `(x-3)x=10`

However 10=5x2 or 2x5, or negatives ones. Let say `x-3=5` then `x=2`

If `x=5` then `x-3=2`

Other way is operate in `x(x-3)=x^2-3x=10` or

`x^2-3x-10=0` using quadratic formula we have

`x=(3+-sqrt(9+40))/2=(3+-7)/2` this arrives to `x=5` and `x=-2`

Lets check the answers `log(-2-3)+log-2=1` reject this solution because nor `log-5` neither `log-2` doesn`t exists

By other hand `log(5-3)+log5=log(2xx5)=log10=1`

So, the only solution is `x=5`