Basics & LogarithmsmediumJEE Main 2022Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2022)

JEE Maths question with a full step-by-step solution.

Question
The sum of all the real roots of the equation (e2x−4)(6e2x−5ex+1)=0(e^{2x}-4)(6e^{2x}-5e^{x}+1)=0 is
Alog⁡e3\log_{e}3
B−log⁡e3-\log_{e}3correct
Clog⁡e6\log_{e}6
D−log⁡e6-\log_{e}6
Solution
Step 1: A product is zero when a factor is zero. So either e2x−4=0e^{2x}-4=0 or 6e2x−5ex+1=06e^{2x}-5e^{x}+1=0. Step 2: First factor: e2x=4⇒ex=2e^{2x}=4\Rightarrow e^{x}=2 (taking positive root since ex>0e^{x}>0). So x=ln⁡2x=\ln 2. Step 3: Second factor: Let u=ex>0u=e^{x}>0:
6u2−5u+1=06u^{2}-5u+1=0
 Step 4: Factor:
(3u−1)(2u−1)=0  ⇒  u=13 or u=12(3u-1)(2u-1)=0\;\Rightarrow\;u=\dfrac{1}{3}\text{ or }u=\dfrac{1}{2}
 Step 5: Solve for xx:
ex=13  ⇒  x=−ln⁡3e^{x}=\dfrac{1}{3}\;\Rightarrow\;x=-\ln 3
ex=12  ⇒  x=−ln⁡2e^{x}=\dfrac{1}{2}\;\Rightarrow\;x=-\ln 2
 Step 6: Sum of all real roots:
ln⁡2+(−ln⁡3)+(−ln⁡2)=−ln⁡3\ln 2+(-\ln 3)+(-\ln 2)=-\ln 3
 Answer: (2) −log⁡e3-\log_{e}3
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Basics & Logarithms · easy
The product of all solutions of the equation e 5( e x) 2 +3 =x 8 , x>0 , is
Basics & Logarithms · easy
The number of real solutions of the equation x (x 2 +3|x|+5|x-1|+6|x-2| )=0 is
Basics & Logarithms · easy
If a+b+c=1 , ab+bc+ca=2 and abc=3 , then the value of a 4 +b 4 +c 4 is equal to
Basics & Logarithms · easy
Sum of all the values of x satisfying the equation 17 11 ( x+11 + x ) = 0 is:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath