ln(1 − x) + ln(4 − x) = ln(10)I would like to know the steps for the answer as well
`ln(1-x) + ln(4-x) = ln 10`
At the left side of the equation, apply the product rule of logarithm which is `log_b m + log_b n = log_b (m*n)` .
`ln [(1-x)(4-x)] = ln 10`
Then, take the anti-logarithm of both sides.
Expand right side.
`x^2 -5x+4 = 10`
Express the equation in a quadratic form `ax^2+bx+c=0` .
Set each factor to zero and solve for x.
`x+1=0` and `x-6=0`
Then, substitute the values of x to the original equation to verify.
`x=-1` , `ln(1-(-1))+ln(4-(-1))= ln 10`
`ln 2+ln 5=ln 10`
`x=6` , `ln(1-6))+ln(4-6)=ln10`
`ln(-5)+ln(-2)=ln10` (Invalid, since in logarithm, the
argument should be positive.)
Hence, the solution to the equation `ln(1-x)+ln(4-x) = ln10` is `x=-1` .