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Learn with Dr. Javed
India
Приєднався 16 лис 2019
This is Dr. Afroz Javed, I completed M.Tech. (Aerospace) from Indian Institute of Technology (IIT) Madras, India, in 1996 and Ph.D. from Indian Institute of Science (IISc), Bengaluru, India, in 2013.In this channel I will be sharing videos for easy and simple understanding of various concepts in the fields of Science, Math, And Engineering.
Also there are lot many things to share like interesting book reviews, comic reviews, some poetry, basics of mathematics, and some interesting scientific experiments observed in daily life...
Also there are lot many things to share like interesting book reviews, comic reviews, some poetry, basics of mathematics, and some interesting scientific experiments observed in daily life...
Only one solution is valid for this logarithmic equation
Hidden quadratic equation behind a logarithmic equation
Only one solution is valid for this logarithmic equation
Use Definition of log
Base of log can not be negative
#logarithmicequation #quadraticequation #learnwithdrjaved
Only one solution is valid for this logarithmic equation
Use Definition of log
Base of log can not be negative
#logarithmicequation #quadraticequation #learnwithdrjaved
Переглядів: 3
Відео
An exponential equation from Oxford Entrance Exam, solved in less than 5 steps
Переглядів 1912 годин тому
An exponential equation from Oxford Entrance Exam, solved in less than 5 steps Natural log Verification is even more satisfying #exponentialequation #oxford #verification
Quick and short solution of a logarithmic equation
Переглядів 3916 годин тому
Quick and short solution of a logarithmic equation Use of division formula of logarithm #logarithmicequation #algebraproblem #quicksolution
How to solve a radical equation using algebraic identity
Переглядів 9121 день тому
How to solve a radical equation using algebraic identity #radicalequation #algebraicidentity #surds
A simple but most basic concept of complex numbers
Переглядів 51Місяць тому
A simple but most basic concept of complex numbers multiply negatives under a surd Basic of complex number operation #complexnumbers #imaginarynumbers #complexanalysis
How to solve a radical equation with complex number
Переглядів 218Місяць тому
How to solve a radical equation with complex number Two negatives can not be multiplied under a surd ua-cam.com/video/xahuv1UMY20/v-deo.html #radicalequation #imaginaryroots #complexnumbers
How to Evaluate Nozzle Exit Mach Number using Microsoft Excel
Переглядів 37Місяць тому
How to Evaluate Nozzle Exit Mach Number using Microsoft Excel Solving an implicit equation How to use Excel for equation solving #rocketscience #equation #exceltips
Thrust of a Rocket Motor and Thrust Coefficient
Переглядів 63Місяць тому
Thrust of a Rocket Motor and Thrust Coefficient #rocketscience #thrust #thrustcoefficient
Solving an exponential equation with logarithmic exponents and trigonometric terms
Переглядів 89Місяць тому
Solving an exponential equation with logarithmic exponents and trigonometric terms Exponential logarithmic trigonometric equation #exponentialequation #logarithmicequation #trigonometricequations
Finding a solution other than the obvious solution for an exponential equation
Переглядів 56Місяць тому
How many solution does this exponential equation has Exponential equation with surds Square root raised to the power square root #exponentialequation #algebra #learnhowtosolvetheexponentialequation
Speed of Exhaust Gases from Rocket Nozzle
Переглядів 772 місяці тому
Speed of Exhaust Gases from Rocket Nozzle Evaluation of Rocket exhaust speed Using Isentropic and ideal gas assumptions #rocketscience #nozzle #exhaust
An exponential Equation with logarithmic exponent
Переглядів 712 місяці тому
An exponential Equation with logarithmic exponent #learnhowtosolvetheexponentialequation #logarithmicequation #algebra
Mass Flow Rate Through a Rocket Nozzle
Переглядів 1183 місяці тому
Derivation of mass flow rate expression through a rocket nozzle using adiabatic expression, energy conservation, and a little of algebraic manipulations. Expression for characteristic Speed or Characteristic Velocity for a Rocket Propellant. #rocketscience #massconservation #nozzledesign
How to Evaluate a logarithmic Expression
Переглядів 993 місяці тому
How to Evaluate a logarithmic Expression Using change of base formula and power law of log
How to evaluate a Logarithmic Expression using given Exponential relation
Переглядів 573 місяці тому
How to evaluate a Logarithmic Expression using given Exponential relation #algebra #howtosolvemathspuzzles #howtosolveolympiadmathproblem
How many solutions this indeterminate equation has?
Переглядів 1158 місяців тому
How many solutions this indeterminate equation has?
An indeterminate exponential equation
Переглядів 1479 місяців тому
An indeterminate exponential equation
How to solve an exponential equation in fractional form
Переглядів 1359 місяців тому
How to solve an exponential equation in fractional form
Can there be a shorter method? - Evaluation of an algebraic expression with the given condition
Переглядів 1049 місяців тому
Can there be a shorter method? - Evaluation of an algebraic expression with the given condition
How to evaluate an arithmetical expression without calculator
Переглядів 1549 місяців тому
How to evaluate an arithmetical expression without calculator
A Nice Olympiad Exponential Problem solved using Heuristic method
Переглядів 2949 місяців тому
A Nice Olympiad Exponential Problem solved using Heuristic method
How to solve this quartic equation for real and imaginary roots
Переглядів 2359 місяців тому
How to solve this quartic equation for real and imaginary roots
An exponential equation with compound surds and quadratic indices with a satisfying solution
Переглядів 28710 місяців тому
An exponential equation with compound surds and quadratic indices with a satisfying solution
How to evaluate the given algebraic expression
Переглядів 17810 місяців тому
How to evaluate the given algebraic expression
A logarithmic equation reduced to fourth order polynomial equation
Переглядів 19210 місяців тому
A logarithmic equation reduced to fourth order polynomial equation
Exponential Logarithmic equation with log in indices
Переглядів 25010 місяців тому
Exponential Logarithmic equation with log in indices
Trignometric function evaluation using given relation
Переглядів 19810 місяців тому
Trignometric function evaluation using given relation
Logarithmic equation with quadratic argument
Переглядів 22410 місяців тому
Logarithmic equation with quadratic argument
An indeterminate equation solved with condition of integer solutions
Переглядів 22210 місяців тому
An indeterminate equation solved with condition of integer solutions
Thanks sir
You are welcome!
👍
Thanks!
🎉🎉🎉 good work
Thanks for the appreciation, Unfortunately very few people know and appreciate these critical details.
🎉🎉🎉 good work
Thanks a lot, you too have a very nice channel doing good work with simple but tricky problems.
Should one solve this equation utilising a quadratic based method, it may be realised that x = (+ or -) (1/2)i Both values may be plugged back in and confirmed to be true.
sqrt (i/2)=(1+i)/2, can be checked by De moivre's formula, so the LHS of the equation becomes ((1+i)/2 )+i((1+i)/2)=i which is not equal to the RHS (1) So the equation has only one solution, that is -i/2.
@@LearnwithDrJaved I'm sorry for taking up your time but this poses a question for me: I believe my misconception was in checking my answer in a calculator as such, sqrt(x) + sqrt(-x), then observing an answer as one for both values of x However, if I instead rewrite sqrt(-x) as i(sqrt(x)) we have: sqrt(x) + i(sqrt(x)) which will only work for the singular value of x being -i/2. My question is, why wouldn't these two equations produce the same result? Furthermore, why would ONLY the one where we use i to remove the negative from the square root be valid? Wolframalpha allows us to see the singular solution for the second equation but both solutions for the first equation, so surely both aren't equal to one another, i.e: sqrt(x) + i(sqrt(x)) does NOT equal sqrt(x) + sqrt(-x) Thank you for your time sir.
x+(-x)+2sqrt(-x.x)=1 after squaring both the sides sqrt (-x^2)=1/2 i . sqrt(x^2)=1/2 sqrt (x^2)=1/2i x = -i/2 (as 1/i = -i) I am perplexed how did you get a positive solution.
I understood the problem, sqrt(-(-i/2)) is not equal to sqrt (i/2), it is equal to i.sqrt(i/2). For more clarity please watch ua-cam.com/video/xahuv1UMY20/v-deo.html
@@LearnwithDrJaved Ah ok thank you
Sir where did the 3/2 come from?
taking log x common on the left hand side, remaining terms add to 3/2
Thanks a lot sir❤
You're welcome. Happy solving:-)
Thank you sir ❤
Happy equation solving, best of luck.
at the very end also error: Gas can be compressed thus inreasing pressure
I could not understand the comment
incorrect formula @t=00:06:24 dp/p
You can check through logarithmic differentiation, this is what you get for dp/p for an isothermal process.
Very nice explanation sir.I am very happy at least few have really tried and made this question easy ...Thanks a lot sir..
This story is just one of the great struggles to get some simple formula we use normally. All the best for your future and studies.
6 is answer Just express in the form of 8 power and 4 power and do in one step
Yes 512=8^3, and 2=√4, will give the result in a single step
How didi u get that a quarter and a half in second step
4V means fourth root, and that is power quarter just V means second root and that is power half
How to solve a square root of 818278 plz plz 🥺🥺
The square root of the given number is an irrational number, you can check with your mobile calculator. The nearest full square number is 818276. And you can evaluate its square root, which comes out to be 904.
Sir i m confused at same question 21/2= 10.5 and 31/29 =1.06 🙏😣🙌😭😭
31/29=1.066896552, this does not pose any problem if you solve long enough you can find the repeating decimal numbers....
Pls explain how you had the answer to be 3/2
OK we got x=y2^2.5 also we have found that y=2^5 this makes y^0.5=2^2.5 we put this value of 2^2.5 in the expression for x and get x=y.y^0.5 x=y^(3/2)
Did it work in the polynomial of degree 8
It MAY work if there are only two terms x^8 and x^4 with a constant.
Technically yes, but you're better off using u-substitution for higher degree polynomials
theres a much faster lest demanding way that takes a few steps, move log4x to the right hand side, multiply log2x (both base 2 and x) by 2 --> log4x^2, then divide the log4x moved to the right hand side by 2 thats been changed into log(4)(4)^2, (4^2/x). Multiply x on both sides and you will get x^3 = 16, take the cube root and you will get the same numerical result
The answer may be correct bu multiplication by 2 part is not clear.
Thats mandarine
Any citrus fruit will do, In fact peels of all the citrus fruits show the same behaviour
yes
To solve any Trigonometric equation, one needs to transform them in these forms!
x[ln(x)]=e [ln(x)][e^[ln(x)]]=e ln(x)=W(e) x=e^W(e) x=e ❤
a very nice alternative solution
⁴√(2x+11)=√5 2x+11=25 2x=14 x=7 ❤
Simplified by squaring both the sides two times!
@@LearnwithDrJaved same thing with raising both sides by the power of 4.
How u got 25 in the second step plz explain 🙏@@ChavoMysterio
(2x+11)^¼=5^½ [(2x+11)^¼]²=(5^½)² (2x+11)^½=5 [(2x+11)^½]²=5² 2x+11=25 That's how you do it.
(2x+11)^¼=5^½ [(2x+11)^¼]⁴=(5^½)⁴ 2x+11=5² 2x+11=25
clear explanation 😁😊
Thanks a lot for your appreciation!
Very good explanation 😊
Thanks a lot for your appreciation!
It stops when all the liquid of the glass is evaporated. So it is not a perpetual motion machine.
It also stops in very cold weather, or humid weather when there is little or no evaporation!
excellent.
Thanks for the appreciation!
❤❤❤❤
Thanks!!!
x=0 and x=3
Yes, these are the two roots of the equation
x=1
x=1 - это решение, но есть и иррациональное решение, как показано на видео
(5-x)^2=x^2-2x+65 x^2-10x+25=x^2-2x+65 25-10x=65-2x -10x+2x=65-25 -8x=40 x=-5
Yes, this is the solution
3^1=1+log3(2^x-7) 3=1+log3(2^x-7) 2=log3(2^x-7) 3^2=2^x-7 9=2^x-7 18=2^x x=2 log 18
The mistake lies in the second line from bottom. as 9+7 = 16 and not 18 as shown in your solution!
Thank you Dr. Javed
You are welcome! And thanks for the appreciation.
❌ = 10 , 1/10
Yes, interestingly both the solutions are reciprocal of each other
+/-3
in addition 4 and 5 are also two of the solutions.
❌ = 1⭕
Nice fonts🙂
25+-4V39.
Вам нужно перепроверить это, единственное решение - x = 100, поскольку база журнала равна 10.
x=4. уравнение сводится к квадратному: m^2-4m-192=0, где m=2^x.
Да, вы также можете использовать эту замену...
x=3.
да, это единственное верное решение
1/4; -1/2. Устно.
Так умно с твоей стороны! Большой.
V19; -V19 -1 .
первый корень - просто наблюдение, второй - хитрый!
x=3; x=9. Спасибо за устное упражнение.
Да, это устно, решение не должно занять более нескольких секунд!
x=5^V2; x=5^(-V2)
Да, это решения!
x=2. Спасибо за устное упражнение.
пожалуйста
-2xy = -96 Xy = 48
Yes, this step also may be included in the solution
x=10. Две секунды устно.
Да, это не должно занять более 5 секунд!
(3^x + 1)(3^x-1) ---------------------------- =8 3^x + 1 3^x -1 =8 3^x =9 3^x =3^2 x= 2
Yes this is a way to get the answer.
(4/9)^-x * (2/3)^3(x-1) = (2/3)^1 (2/3)^-2x * (2/3)^3x-3 = (2/3)^1 (2/3)^-2x+3x-3 = (2/3)^1 -2x+3x-3=1 x-3=1 x=4
A great quick solution, thanks!
x=2. Использована формула "разность квадратов".
Answer + 1/3 , -1/3 , + 1 and -1 let x^-2 = n, then 9 + n^2 = 10n n^2 -10n -9 =0 (n-9) (n-1) =0 n=9 and n=1 hence 9 = x^ -2 and 1= x^-2 9 = 1/x^2 and 1 = 1/x^2 1/9 =x^2 1/1 = x^2 + or -1/3 = x +1 or -1 = x
Thanks for providing a solution with the method of substitution
1/3 and 1
-1/3 and -1 are also solutions.