$x$ is a literal number and it is involved in logarithmic and algebraic systems to form an equation.
$\dfrac{\log_{2} {(9-2^{\displaystyle x})}}{3-x} \,=\, 1$
It is required to find the solution of this equation to know the value of the $x$.
Apply cross multiplication rule to express the equation in simple form.
$\implies$ $\log_{2} {(9-2^{\displaystyle x})}$ $\,=\,$ $1 \times (3-x)$
$\implies$ $\log_{2} {(9-2^{\displaystyle x})}$ $\,=\,$ $3-x$
Eliminate logarithmic form from the equation and it can be done by using the relation between logarithms and exponential notation.
$\implies$ $9-2^{\displaystyle x} \,=\, 2^{\displaystyle 3-x}$
$\implies$ $9-2^{\displaystyle x} \,=\, 2^{\displaystyle 3} \times 2^{\displaystyle -x}$
$\implies$ $9-2^{\displaystyle x} \,=\, 8 \times 2^{\displaystyle -x}$
$\implies$ $9-2^{\displaystyle x} \,=\, \dfrac{8}{2^{\displaystyle x}}$
$\implies$ $2^{\displaystyle x}(9-2^{\displaystyle x}) \,=\, 8$
$\implies$ $9(2^{\displaystyle x})-{(2^{\displaystyle x})}^2 \,=\, 8$
$\implies$ $0 \,=\, {(2^{\displaystyle x})}^2 -9(2^{\displaystyle x}) + 8$
$\implies$ ${(2^{\displaystyle x})}^2 -9(2^{\displaystyle x}) + 8 \,=\, 0$
The equation is in the form of quadratic equation. It can be solved by using the methods of the solving quadratic equations. Take $v = 2^{\displaystyle x}$ to avoid confusion in solving the quadratic equation.
$\implies$ $v^2-9v+8 \,=\, 0$
The quadratic equation can be solved by using the factoring method.
$\implies$ $v^2-8v-v+8 \,=\, 0$
$\implies$ $v(v-8)-1(v-8) \,=\, 0$
$\implies$ $(v-1)(v-8) \,=\, 0$
Therefore, $v \,=\, 1$ and $v \,=\, 8$
As per our assumption, the value of literal $v$ is $2^{\displaystyle x}$. So, replace it to obtain the value of the $x$.
$2^{\displaystyle x} = 1$
$\implies 2^{\displaystyle x} = 2^0$
$\implies x = 0$
$2^{\displaystyle x} = 8$
$\implies 2^{\displaystyle x} = 2^3$
$\implies x = 3$
The two cases have given two solutions to the logarithmic equation. Therefore, the values of $x$ are $0$ and $3$.
Now, check the logarithmic equation at $x$ is equal to $0$ and also $x$ is equal to $3$.
$\dfrac{\log_{2} {(9-2^{\displaystyle 0})}}{3-0}$
$= \dfrac{\log_{2} {(9-1)}}{3}$
$= \dfrac{\log_{2} 8}{3}$
$= \dfrac{\log_{2} 2^3}{3}$
$= \dfrac{3 \log_{2} 2}{3}$
$= \require{cancel} \dfrac{\cancel{3} \log_{2} 2}{\cancel{3}}$
$= \log_{2} 2$
Apply, the logarithm of base rule to obtain the value of the expression.
$= 1$
The value of the left hand side expression is equal to $1$ and it is the value of the right hand side of the equation. Hence, the value of $x$ equals to $0$ is true solution of the equation.
$\dfrac{\log_{2} {(9-2^{\displaystyle 3})}}{3-3}$
$= \dfrac{\log_{2} {(9-8)}}{0}$
$= \dfrac{\log_{2} {(1)}}{0}$
$= \dfrac{0}{0}$
Therefore, the value of left hand side expression is indeterminate at $x$ is equal to $3$. Hence, $x \ne 3$ but $x = 0$ is only the solution of the logarithmic equation and it is required solution for this logarithmic problem mathematically.
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