#### Answer

$x=25$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $ \log x+\log (x-21)=\log100 ,$ use the properties of logarithms to simplify the left-hand expression. Then drop the logarithm on both sides. Use the properties of equality and concepts on solving quadratic equations. Finally, do checking of the solutions with the original equation.
$\bf{\text{Solution Details:}}$
Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log [x(x-21)]=\log100 .\end{array}
Since the logarithmic expressions on both sides have the same base, then the logarithm may be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} x(x-21)=100 .\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x(x)+x(-21)=100 \\\\ x^2-21x=100 .\end{array}
Using the properties of equality, in the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x^2-21x-100=0 .\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is
\begin{array}{l}\require{cancel}
(x-25)(x+4)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x-25=0
\\\\\text{OR}\\\\
x+4=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x-25=0
\\\\
x=25
\\\\\text{OR}\\\\
x+4=0
\\\\
x=-4
.\end{array}
If $
x=-4
,$ the part of the given expression, $
\log x
,$ becomes $
\log (-4)
.$ This is not allowed since $
\log x$ is defined only for nonnegative values of $x.$ Hence, only $
x=25
$ satisfies the original equation.