Math Doubts

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

$x$ 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 $x$.


Cross multiplication rule

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

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

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


Transformation of equation in exponential form

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

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

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

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

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

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

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

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

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


Solve Quadratic equation

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 $v \,=\, 8$


Evaluating solution

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 $3$.


Verifying the Roots

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.

Math Questions

The math problems with solutions to learn how to solve a problem.

Learn solutions

Math Worksheets

The math worksheets with answers for your practice with examples.

Practice now

Math Videos

The math videos tutorials with visual graphics to learn every concept.

Watch now

Subscribe us

Get the latest math updates from the Math Doubts by subscribing us.

Learn more

Math Doubts

A free math education service for students to learn every math concept easily, for teachers to teach mathematics understandably and for mathematicians to share their maths researching projects.

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved