$x$ 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 $x$.

01

Apply cross multiplication rule to express the equation in simple form.

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

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

02

Eliminate logarithmic form from the equation and it can be done by using the relation between logarithms and exponential notation.

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

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

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

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

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

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

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

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

03

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 $v \,=\, 8$

04

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 $3$.

05

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.

List of most recently solved mathematics problems.

Jul 04, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \,\to\, \tan^{-1}{3}} \normalsize {\dfrac{\tan^2{x}-2\tan{x}-3}{\tan^2{x}-4\tan{x}+3}}$

Jun 23, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \to 0} \normalsize \dfrac{e^{x^2}-\cos{x}}{x^2}$

Jun 22, 2018

Integral Calculus

Evaluate $\displaystyle \int \dfrac{1+\cos{4x}}{\cot{x}-\tan{x}} dx$

Jun 21, 2018

Limit

Evaluate $\displaystyle \large \lim_{x \to \infty} \normalsize {\sqrt{x^2+x+1}-\sqrt{x^2+1}}$

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.