• Algebra
• Trigonometry
• Geometry
• Calculus

Solve $\dfrac{\log_{2} {(9-2^x)}}{3-x} \,=\, 1$ and find the value of $x$

$x$ 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 $x$.

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.