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

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lemjay | High School Teacher | (Level 2) Senior Educator

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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)`


Expand right side.

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

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


Then. factor.


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`



`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` .

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