Question

How do you solve `Logx + Log(x+9)=1`?

Answer 1

See below

Explanation:

Our goal is to have an expresion like this `logA=logB` due to one to one aplication of logarithm, we get `A=B`. We have to apply the rules of logarithm for product, quotient and power and use the fact `log10=1`

`logx+log(x+9)=1`
`logx(x+9)=log 10`

Now, we know that `x(x+9)=10`
`x^2+9x-10=0` using quadratic formula we get

`x=-10 and x=1` reject negative because there is no logarithm and the only solution is `x=1`. Lets check it

`log1+log(1+9)=0+1=1`

Answer 2

`x=1`

Explanation:

`logx+log(x+9)=1`

`=>log(x(x+9))=1`

i.e. `x(x+9)=10`

or `x^2+9x-10=0`

or `x^2-x+10x-10=0`

or `x(x-1)+10(x-1)=0`

or `(x-1)(x+10)=0`

so either `x-1=0` i.e. `x=1`

or `x+10=0` i.e. `x=-10`

but we cannot have `x<=-9` as `x+9>0`

Hence only solution is `x=1`