Algebra Adventure: Conquering Exponential Equations with Ease!

Algebra Adventure: Conquering Exponential Equations with Ease!

In this exciting algebraic adventure, we take on the challenge of solving exponential equations and provide you with the tools and techniques to conquer them with ease. Join us as we unravel the mysteries of exponential equations and guide you through a step-by-step process to solve them confidently. Whether you're a student struggling with algebra or someone looking to sharpen your problem-solving skills, this video is packed with valuable insights and strategies. Discover the power of logarithms, exponential properties, and other algebraic concepts that will help you crack even the most complex exponential equations. Get ready to embark on this educational journey and witness your confidence soar as you overcome these algebraic challenges. Watch now and become an exponential equation-solving master!

Topics covered:
Algebra challenging problem
Exponential Equation
Lambert W function
Graphs
Math Olympiad preparation
Mathematics Challenging equations
Problem solving

#AlgebraAdventure #ExponentialEquations #MathMadeEasy #MathTips #ProblemSolving

8 Key moments of this video :
0:00 Introduction
0:48 Exponent rules
1:33 Substitution
2:32 Algebraic manipulation
2:57 Lambert W Function
5:40 Solving for x
7:01 Graphical conclusions
7:17 Solution using Calculus

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Thanks for watching !!
@infyGyan ​

[namangarg]

I have prepared for jee advanced and this ques is just cup of tea for me i used first algebra to simplify by dividing by 2 then use graph of both to get know how many solution that is unique because they cut for unique value of x then analyse that LH FUNC IS INC AND RH FUNC IS DECREASING. Then just limit the x values between 0 to 11 then just put integer values in equation then by hit and trial x=3 is solution because if x is not integrr then we have to use another simplifying method to get the solition

[nasrullahhusnan]

We can solve the problem without the use of Lambert W function.
The given equation 2^(x+1)+2x=22
Divide by 2: 2^x+x=11 (1)
Take modulo 2: note that modulo 2 of the first term in LHS is 0. Hence mod(x, 2)=1, meaning that x is odd number.
Take modulo 3:
• mod(2^x, 3)=[mod(2, 3)]^x
=(-1)^x
=-1 as x is odd
•mod(11, 3)=-1
Therefore -1+mod(x, 3)=-1
mod(x, 3)=0 --> 3|x
Thus x=3k where k is any integer. Plug it to (1): 2^(3k)+3k=11
8^k+3k=11
It is clear that k=1 and consequently x=3.

[chrisong]

Hi, anyone can tell me how to get ((2^11)×ln2)÷ln2=8?
Kindly reply ASAP. Thanks.