Explain how you would solve the following equation:

$\mathrm{ln}\left(x\right)+\mathrm{ln}(x-5)=\mathrm{ln}(21-x)$

Describe why you may only choose some of the possible roots of any polynomial you reduce the problem to

Step 1

By using the property of logarithms, simplify the LHS .

$\mathrm{ln}\left(a\right)+\mathrm{ln}\left(b\right)=\mathrm{ln}\left(ab\right)$

$\text{Simplify the LHS}$

$\mathrm{ln}\left(x\right)+\mathrm{ln}(x-5)=\mathrm{ln}\left(x(x-5)\right)$

$=\mathrm{ln}({x}^{2}-5x)$

Step 2

Since there is log term on both the sides of the equation, we can take antilog on both sides, simplify and solve the quadratic for x

$\mathrm{ln}({x}^{2}-5x)=\mathrm{ln}(21-x)$

$e}^{\mathrm{ln}({x}^{2}-5x)}={e}^{\mathrm{ln}(21-x)$

${x}^{2}-5x=x-21$

${x}^{2}-5x-x-21=0$

${x}^{2}-4x-21=0$

${x}^{2}-7x+3x-21=0$

$x(x-7)+3(x-7)=0$

$(x-7)(x+3)=0$

$\Rightarrow x=7,x=-3$

Step 3

There are two solutions possible for value of x, one positive the other negative. However notice that in the given question there is a term ln(x). But since log of a number is defined only for x>0, we discard x=-3.

Hence the answer is x=7.