I came here to say this. I also used exponent and log properties to eliminate the log notation on the left side. Math TH-camrs are super weird sometimes.
I love how even though sometimes there may be an easier way to solve a problem you always manage to show a unique solution. You allow us to relearn or review old rules we may have forgotten and you let us see your mental flow of the problems at hand! You inspire me to be a better teacher every day!
considering x > 0 and not equal to one, we can proceed by using the properties of exponents on the LHS and by property of logarithms the LHS simplifies to 16x = 4x^2 => 4x = x^2 ===> x = 0,4 but we can't take 0 and we assumed x not equal to 0 before solving this hence x must be equal to 4
x=0 and x=4 i think these are the answers lemme check the vid now edit: checked it and i now know that x cannot be zero just split x^(logx(16) + 1) into (x^logx(16))*(x^1) and since the logx cancels the x it becomes 16 * x or 16x 16x=4x^2 4x=x^2 x^2-4x=0 x(x - 4) = 0 x = 0 and x = 4 of which now i know x=0 is not an answer as log base 0 is not possible
can't you just use exponent properties to turn the LHS into x^1 * x^log_x(16), which is just 16x?
Damn is that not what he does? I haven’t started watching yet because I like to solve these before watching him walk through it and that’s what I did
but if you do that you get x=0 and x=4 but x can't equal 0 because you can't take log base 0 of a number.
@@redroach401 in the solution finding process, when we do “x^log_x(16)=16” we are quietly assuming that x != 0,1
@@frimi8593 Oh ok thank you
I came here to say this. I also used exponent and log properties to eliminate the log notation on the left side.
Math TH-camrs are super weird sometimes.
I love how even though sometimes there may be an easier way to solve a problem you always manage to show a unique solution. You allow us to relearn or review old rules we may have forgotten and you let us see your mental flow of the problems at hand! You inspire me to be a better teacher every day!
This is one of the best math channel on TH-cam!
I love your hand writing too
Thank you so much, Sir. Your dedication is amazing. Please continue and never stop, I love your slogan as well!
"I know this is easy, but I just missed it" is the most relatable thing
It Never gets old.
Thank you for reminding us the basics sir.
Maravillosa resolución , eres excelente profesor, saludos y bendiciones
One of the best teachers! Its always great to see your videos! Thank you very much!
Thank you! 😃
1st row: our equation.
2nd row: 16x = 4x²
3rd row: 4x² - 16x = 0
4x(x-4) = 0
The X can't be 0, so x=4.
why x cannot equal to 0?
@@radzelimohdramli4360, according to the definition of the logarithm, the logarithm base cannot be equal to 0
Why can't x be zero? Zero as a base is a little boring, but it is allowed, right?
Okay, the Internet says it's not allowed, but all I could find were some kind of hand-waving type videos to explain why.
Needs more thought 🤔
Great my sir...long live
Nice and easy
considering x > 0 and not equal to one, we can proceed by using the properties of exponents on the LHS and by property of logarithms the LHS simplifies to 16x = 4x^2 => 4x = x^2 ===> x = 0,4 but we can't take 0 and we assumed x not equal to 0 before solving this hence x must be equal to 4
i believe i have an easier solution
x . x^log_x(16)=4x^2
canceling out x s
x^log_x(16)=4x
x^log_x(16) = 16 so
16=4x
4=x
great video. if i had had this in HS math class, i would have run away from home and joined a circus
Another cool video and math development, although it could have been solved in four or so steps by observing that
x^(1 + logx(16)) = x*x^(logx(16)).
i just did:
x^(1+log(x, 16)) = 4x^2 ;; equation
x^1 * x^log(x, 16) = 4x^2 ;; expanding
16x = 4x^2 ;; simplifying
4x=x^2
(x^2)/x=4 ;; you can already solve it here
(x^2)*(x^-1)=4
x=4 ;; answer, using exponent rules
Teacher: "Those who stop learning stop living"
Students: scared for their lives
Math ❤
Nice problem great video! May I say that the camera was losing focus quite a bit which was annoying.
I solved it but when I got 4 I had a heart attack cause it was an integer
😀
x=0 and x=4 i think these are the answers lemme check the vid now
edit: checked it and i now know that x cannot be zero
just split x^(logx(16) + 1) into (x^logx(16))*(x^1) and since the logx cancels the x it becomes 16 * x or 16x
16x=4x^2
4x=x^2
x^2-4x=0
x(x - 4) = 0
x = 0 and x = 4
of which now i know x=0 is not an answer as log base 0 is not possible
u can its much simpler without taking log of both sides
Yeah when you cancel things out you sometimes have to carry extra conditions like x ≠ 0 forward. Same when you cancel x/x
downside of log can't be zero.
@@pearlwoodsword5965 Yeah you do get the correct answer then
4
is x=4 the only answer? what about x=0?
Base cannot be 0
How are you? I have one question
x=0,4 , not only 4
Base of log can’t be 0
Just cancel the x on both sides in the beginning. X cannot be 0 since it makes log undefined. Stop beating around the bush for view-time.