Solving a logarithmic equation

[@Forgettings]

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!

[@SuperTommox]

This is one of the best math channel on KZbin! I love your hand writing too

[@cliveanawana5289]

Thanks for this video. btw I was able to solve it without applying the log properties. 1. Divide both sides by X^2. The RHS becomes 4 and the LHS becomes X^(log.base X of 16) minus 1 2. Split the LHS into two as X^(log.base X of 16) divided by X^(log.base X of X, which replaces 1 in the previous line) 3. Xs cancel each other which gives an expression of 16/x = 4 4. Simple algebra knowledge gives x = 4

[@pook_7458]

can't you just use exponent properties to turn the LHS into x^1 * x^log_x(16), which is just 16x?

[@frimi8593]

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

[@redroach401]

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.

[@frimi8593]

@@redroach401 in the solution finding process, when we do “x^log_x(16)=16” we are quietly assuming that x != 0,1

[@redroach401]

@@frimi8593 Oh ok thank you

[@emmeeemm]

I came here to say this. I also used exponent and log properties to eliminate the log notation on the left side. Math KZbinrs are super weird sometimes.

[@V2II-r1f]

Thank you so much, Sir. Your dedication is amazing. Please continue and never stop, I love your slogan as well!

[@nathanisbored]

"I know this is easy, but I just missed it" is the most relatable thing

[@mkmathstutorials6645]

It Never gets old. Thank you for reminding us the basics sir.

[@2106522]

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.

[@radzelimohdramli4360]

why x cannot equal to 0?

[@2106522]

​@@radzelimohdramli4360, according to the definition of the logarithm, the logarithm base cannot be equal to 0

[@lornacy]

Why can't x be zero? Zero as a base is a little boring, but it is allowed, right?

[@lornacy]

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 🤔

[@cesarbasilio2804]

One of the best teachers! Its always great to see your videos! Thank you very much!

[@PrimeNewtons]

Thank you! 😃

[@makhosimayisa5800]

Nice and easy

[@luisalfredonarvaeznarvaez5125]

Maravillosa resolución , eres excelente profesor, saludos y bendiciones

[@biswambarpanda4468]

Great my sir...long live

[@Harrykesh630]

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

[@lukasjetu9776]

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

[@faustobarbuto]

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)).

[@FreeAcademyForMath]

If you separate the lhs to a product where both bases are x, then things come together pretty quickly.

[@ilkayaktas7699]

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

[@heythere9380]

great video. if i had had this in HS math class, i would have run away from home and joined a circus

[@amtep]

Teacher: "Those who stop learning stop living" Students: scared for their lives

[@issazeko6133]

Nice problem great video! May I say that the camera was losing focus quite a bit which was annoying.

[@radzelimohdramli4360]

is x=4 the only answer? what about x=0?

[@PrimeNewtons]

Base cannot be 0

[@aMyst_1]

I solved it but when I got 4 I had a heart attack cause it was an integer

[@PrimeNewtons]

😀

[@mrmimi807]

4

[@samkid123]

Math ❤

[@HeydaraAbdurehman]

How are you? I have one question

[@cmagicalex]

x=0,4 , not only 4

[@chonkeboi]

Base of log can’t be 0

[@hamzaiqbal7178]

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

[@pearlwoodsword5965]

u can its much simpler without taking log of both sides

[@amtep]

Yeah when you cancel things out you sometimes have to carry extra conditions like x ≠ 0 forward. Same when you cancel x/x

[@saropasha]

downside of log can't be zero.

[@hamzaiqbal7178]

@@pearlwoodsword5965 Yeah you do get the correct answer then

[@raja2850]

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.