Question

How do you solve `log(x-5) + log(x-2)=1`?

Answer 1

`x=7`

Explanation:

`log(x-5)+log(x-2)=1`
`color(white)("XXXXXXXX")`since `log(a*b) = log(a)+log(b)`
`rArrlog( (x-5)(x-2) ) =1`
`color(white)("XXXXXXXX")`expanding the multiplication
`log( x^2-7x+10 ) =1`
`color(white)("XXXXXXXX")`despite other comments the default base for `log` is `10`
`color(white)("XXXXXXXX")`so using the above as exponents of `10`
`10^(log(2x^2-7x+10))=10^1`
`rArrx^2 - 7x +10 = 10`
`color(white)("XXXXXXXX")`subtracting `10` from both sides
`x^2-7x=0`
`color(white)("XXXXXXXX")`factoring
`x(x-7)=0`

i.e. `x=0` or `x=7`

However `log(x-5)` and `log(x-2)` are undefined if `x=0`;
so `x=0` is an extraneous solution
and only `x=7` is valid

Answer 2

`x=7`

Explanation:

As no base is given, it is assumed to be `10`
Natural logs are commonly denoted by ln.

`log(x-5) +log(x-2) =color(blue)(1)`

In an expression or equation, the terms must all be in the same form - either all logs or all numbers.

`log(x-5) +log(x-2) =color(blue)(log10)" "larr(color(blue)(log_10 10 hArr 1))`

`color(white)(xxxxxxxxxxx)`Apply the law: `" "loga + logb hArr log(ab)`

`log((x-5)(x-2)) =log10`

`color(white)(xxxxxxxxxxx)`Apply the law: `" "log a = logb hArr a=b`

`:.(x-5)(x-2) = 10color(white)(xxxxxxxxxxx)`'drop' the logs

`x^2-7x+10 = 10" "larr` solve the quadratic equation

`x^2-7x=0" "larr` factorise

`x(x-7)=0`

`x =0 or x=7`

However, `x=0` is an extraneous solution, and not valid in this equation.

`x=7`