Solve $\dfrac{\log_{2}{(9-2^x)}}{3-x}$ $\,=\,$ $1$

Analyze the equation for Simplification

The expression on the left-hand side of the equation is in fraction form and the expression on the right-hand side of the equation is a number. First of all, it should be simplified to clear the route for solving this log equation and it can be done by writing the expressions on both sides in same form.

$\implies$ $\dfrac{\log_{2}{(9-2^{\displaystyle x})}}{3-x}$ $\,=\,$ $\dfrac{1}{1}$

Now, the above equation in fraction form can be simplified by using the cross multiply rule.

$\implies$ $1 \times \log_{2}{(9-2^{\displaystyle x})}$ $\,=\,$ $1 \times (3-x)$

$\,\,\,\,\therefore\,\,\,\,$ $\log_{2}{(9-2^{\displaystyle x})}$ $\,=\,$ $3-x$

Simplify the Exponential equation

The given logarithmic equation is converted as an exponential form equation in the above step. Now, it is time to simplify the exponential equation and it helps us to prepare the equation for solving the value of $x$.

$\implies$ $9-2^{\displaystyle x}$ $\,=\,$ $2^{3- \displaystyle x}$

$\implies$ $9-2^{\displaystyle x}$ $\,=\,$ $2^3 \times 2^{\displaystyle -x}$

$\implies$ $9-2^{\displaystyle x}$ $\,=\,$ $8 \times 2^{\displaystyle -x}$

$\implies$ $9-2^{\displaystyle x}$ $\,=\,$ $8 \times \dfrac{1}{2^{\displaystyle x}}$

$\implies$ $9-2^{\displaystyle x}$ $\,=\,$ $\dfrac{8}{1} \times \dfrac{1}{2^{\displaystyle x}}$

$\implies$ $9-2^{\displaystyle x}$ $\,=\,$ $\dfrac{8 \times 1}{1 \times 2^{\displaystyle x}}$

$\implies$ $9-2^{\displaystyle x}$ $\,=\,$ $\dfrac{8}{2^{\displaystyle x}}$

$\implies$ $2^{\displaystyle x}(9-2^{\displaystyle x}) \,=\, 8$

$\implies$ $9(2^{\displaystyle x})-{(2^{\displaystyle x})}^2 \,=\, 8$

$\implies$ $0 \,=\, {(2^{\displaystyle x})}^2 -9(2^{\displaystyle x}) + 8$

$\implies$ ${(2^{\displaystyle x})}^2 -9(2^{\displaystyle x}) + 8 \,=\, 0$

Solve the Equation in Quadratic form

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 $x$.

Case: 1

$2^{\displaystyle x} = 1$

$\implies 2^{\displaystyle x} = 2^0$

$\implies x = 0$

Case: 2

$2^{\displaystyle x} = 8$

$\implies 2^{\displaystyle x} = 2^3$

$\implies x = 3$

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 $3$.

Substitute x = 0

$\dfrac{\log_{2} {(9-2^{\displaystyle 0})}}{3-0}$

$= \dfrac{\log_{2} {(9-1)}}{3}$

$= \dfrac{\log_{2} 8}{3}$

$= \dfrac{\log_{2} 2^3}{3}$

$= \dfrac{3 \log_{2} 2}{3}$

$= \require{cancel} \dfrac{\cancel{3} \log_{2} 2}{\cancel{3}}$

$= \log_{2} 2$

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.

Substitute x = 3

$\dfrac{\log_{2} {(9-2^{\displaystyle 3})}}{3-3}$

$= \dfrac{\log_{2} {(9-8)}}{0}$

$= \dfrac{\log_{2} {(1)}}{0}$

$= \dfrac{0}{0}$

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.