Question

The number of real roots of the equation ${5+}\left| {{2}^{x}}-1 \right|={{2}^{x}}\left( {{2}^{x}}-2 \right)$ is
A) 2
B) 3
C) 4
D) 1

Answer 1

In this we will find the roots of $\text{5+}\left| {{2}^{x}}-1 \right|={{2}^{x}}\left( {{2}^{x}}-2 \right).$ By consider first $\left| {{2}^{x}}-1 \right|$ positive and then converting the given equation it into quadratic equation and also by taking $\left| {{2}^{x}}-1 \right|$ negative and then converting the given equation into quadratic equation. By find the roots of converted quadratic we will find the roots of equation $\text{5+}\left| {{2}^{x}}-1 \right|={{2}^{x}}\left( {{2}^{x}}-2 \right).$
Consider a quadratic equation $a{{x}^{2}}+bx+c=0$. To find the roots of a quadratic equation we will split the coefficient of x i.e. b its sum will be the product a x c the by taking the common term out we will get factorisation of quadratic equation i.e. we will get roots of quadratic equation.

Complete step-by-step answer:
The given equation is,
$\text{5+}\left| {{2}^{x}}-1 \right|={{2}^{x}}\left( {{2}^{x}}-2 \right).....(1)$
When $\left| {{2}^{x}}-1 \right|$ positive
$\left| {{2}^{x}}-1 \right|={{2}^{x}}-1$
Equation (1) becomes
$\text{5+}{{2}^{x}}-1={{2}^{x}}\left( {{2}^{x}}-2 \right)$
$\text{4+}{{2}^{x}}={{2}^{2x}}-2\cdot {{2}^{x}}$
${{2}^{2x}}-2\cdot {{2}^{x}}-{{2}^{x}}-4=0$
${{2}^{2x}}-3\cdot {{2}^{x}}-4=0$
Put ${{2}^{x}}=t$
${{t}^{2}}-3t-4=0$
Which is the quadratic equation we can solve by factorisation method i.e. we split the coefficient of t such that their product is the last constant term.

& {{t}^{2}}-4t+t-4=0 \\\ & t\left( t-4 \right)+1\left( t-4 \right)=0 \\\ & \left( t+1 \right)\left( t-4 \right)=0 \\\ \end{aligned}$$ $$\left( t+1 \right)=0\text{ or }\left( t-4 \right)=0$$ t = -1 or 4 Since, $$t={{2}^{x}}$$ $${{2}^{x}}=-1\text{ or }{{2}^{x}}=4$$ By taking logarithm, we get $$\log {{2}^{x}}=\log \left( -1 \right)\text{ or log}{{2}^{x}}=\log 4$$ Since logarithm of negative numbers is not possible. $$\log {{2}^{x}}\ne \log \left( -1 \right)$$ $$\text{log}{{2}^{x}}=\log 4=\log {{2}^{2}}$$ $$\text{xlog}2=2\log 2$$ By cancelling log2 on both sides, we get x=2. When $$\left| {{2}^{x}}-1 \right|$$ negative $$\left| {{2}^{x}}-1 \right|=-\left( {{2}^{x}}-1 \right)=-{{2}^{x}}+1$$ Equation (1) becomes $$\text{5-}{{2}^{x}}+1={{2}^{x}}\left( {{2}^{x}}-2 \right)$$ $$\text{6-}{{2}^{x}}={{2}^{2x}}-2\cdot {{2}^{x}}$$ $${{2}^{2x}}-2\cdot {{2}^{x}}+{{2}^{x}}-6=0$$ $${{2}^{2x}}-{{2}^{x}}-6=0$$ Put $${{2}^{x}}=t$$ $${{t}^{2}}-t-6=0$$ Which is the quadratic equation we can solve by factorisation method i.e. we split the coefficient of t such that their product is the last constant term. $$\begin{aligned} & {{t}^{2}}-3t+2t-6=0 \\\ & t\left( t-3 \right)+2\left( t-3 \right)=0 \\\ & \left( t+2 \right)\left( t-3 \right)=0 \\\ \end{aligned}$$ $$\left( t+2 \right)=0\text{ or }\left( t-3 \right)=0$$ t = -2 or 3 Since, $$t={{2}^{x}}$$ $${{2}^{x}}=-2\text{ or }{{2}^{x}}=3$$ By taking logarithm, we get $$\log {{2}^{x}}=\log \left( -2 \right)\text{ or log}{{2}^{x}}=\log 3$$ Since logarithm of negative numbers is not possible. $$\log {{2}^{x}}\ne \log \left( -2 \right)$$ $$\text{log}{{2}^{x}}=\log 3$$ $$\text{xlog}2=\log 3$$ $$\text{x}=\dfrac{\log 3}{\text{log}2}$$ Hence there are two real root of equation $$\text{5+}\left| {{2}^{x}}-1 \right|={{2}^{x}}\left( {{2}^{x}}-2 \right).$$ **So, the correct answer is “Option A”.** **Note:** In this problem, students should note that logarithm of negative numbers is not possible. Alternative method to find roots of quadratic equations. Consider the quadratic equation $$a{{x}^{2}}+bx+c=0$$. Then roots of the equation can be find by following formula: $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.