# ln(1 − x) + ln(4 − x) = ln(10)I would like to know the steps for the answer as well

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`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.

`e^(ln[(1-x)(4-x)])= e^(ln10)`

`(1-x)(4-x)=10`

Expand right side.

`x^2 -5x+4 = 10`

Express the equation in a quadratic form `ax^2+bx+c=0` .

`x^2-5x-6=0`

Then. factor.

`(x+1)(x-6)=0`

Set each factor to zero and solve for x.

`x+1=0` and `x-6=0`

`x=-1` `x=6`

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`

`ln(2*5)=10`

`ln10=ln10`

`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` .**