Logarithms Review (1 of 4: Using Log Laws to solve Log Equations)

[@bananaking3737]

Thanks so much! I've been binge-watching your videos, probably 25 of them in 3 days and have taken down around 10 pages of notes which I will study off! You are so generous and kind, and an amazing teacher! :)

[@username_exe]

Why aren't all mathematics teachers like him. I mean, after I watched this video, I'm now able to do the sums in my book whereas these are the same things that I found very difficult to understand in class or tuition for a whole year!!! He's just so brilliant!!!!

[@adilabdu18]

I wish all the teachers were like you !!!

[@srirammk2609]

OMG, he doesn't monetize his video!! LEGEND SIR!

[@Liquid3_]

what that mean?

[@sakshamsinha8093]

@@Liquid3_ making money by adding ads to a video

[@tienganhvacuocsong7943]

The more I watch your videos, the more I want to appreciate you a lot, sir. Thank you very much for your help.

[@JohnDoe-hw6sq]

if all the teachers tried so hard to present a lesson like this .. I conclude that there are no bad students, there are only bad teachers

[@vinceallenmeneses5883]

bravo sir. ive watched 10 of your vids sraight. ive graduated 12yrs ago. i still love math. 😙

[@okccitizen4400]

Bravo 👏🏼

[@elv4evapersonal447]

U have my respect. Same xDD

[@unnatshaneshwar6477]

same but i haven't yet graduated though i hated maths at a point but after actually understanding it i'm in love with it (thanks to youtube)

[@coleabrahams9331]

I loved his way of teaching accompanied by the review questions. Thanks so much

[@nakitumizajashi4047]

I do not consider myself smart, but those questions that students were asking are really, really easy to answer. Somebody was not paying attention...

[@Shpoopdoop]

I understood what he was saying but then again he does go a bit faster than others. I think no question is a stupid one if one's acquiring knowledge from it.

[@davidvanbrackle589]

where have teachers like this been all my life?!

[@peniaszulu9356]

great teaching i never thought i could understand

[@admiraljank697]

epic chair stumble at 4:43

[@danielkivari9329]

I love your videos, but I wished that you titled this one, "Evaluating Logarithmic Expressions"!

[@adamthethird4753]

Anyone else actually do the problems before he explained them? I had trouble reading number 2 so here is an easier to read format. 1. 5Log8 2 + (1/2)log8 4 2. log3 270 -(log3 2 + log3 5) 3. loga a^2 + 3loga a 4. loga sqrt(x) - log a (1/x)

[@HaydenNK3]

Two months later, I know I reply a bit late but I want to say thank you (even if I personally didn't really need these "easier reading format" of the problems) It's always appreciable to see someone who cares about others ^^

[@johndubose1401]

Love your presentations. Help to better understand logs! Thanks!

[@hassanibrahim5032]

Mr Woo you are Great love how you explain concepts simply and clearly + your high energy God Bless you :)

[@GhostBrew]

I know this is a old video, but if you do happen to see the comment Professor Woo, what if you choose to use exponents for the last example no.4, then it would be, log base a x to the power of 1/2 - log base a x to the power of -1, which is, log base a x to the power of ((1/2)/-1) which then simplifies to, log base a x to the power -(1/2) which = -log base a square root of x, now I don't get what you get and therefore my question is what makes my method wrong, I would love your reason so that I can see what I am missing......

[@sanjogkabare4737]

The problem with your logic comes in step 2. Using the quotient rule, the equation simplifies to log base a of, x raised to (1/2) divided by x raised to negative 1. Using law of indices, the powers of x get subtracted (not divided). 1/2-(-1)=3/2 So the equation simplifies to log base a of x raised to 3/2, agreeing with his method. Hope you got your mistake.

[@RazorM97]

this was so useful

[@lacimoore2618]

On question #4 when you added the logs, you added the fraction but did not multiply the x's within the log. Where they suppose to be multiplies based on the Law of adding Log?

[@jomialsipi]

log(x) + log(x) = log(x^2) = 2 log(x). Adding the coefficients is equivalent to multiplying, but since you already have coefficients it's easier to simply add them together.

[@jyotialpula1705]

I had doubt in 4th question 2nd step. how did you get 3by2

[@jana1476]

4:43 in slo mo made my school career!

[@Emilyrosepears]

I got stuck doing these questions when it asked me to find x. If you're working with log base e or ln do you always need to check your solutions?? I thought x couldn't be negative but sometimes it seems it can be and you have to check.. I guess my question is how do you know when to check and when is it okay to simply say x>0

[@MrSG-007]

can u pls. put videos covering number theory in details.

[@silveriosierra1258]

a delight to see smart students....

[@eamonburns9597]

At 8:33, wouldn't it be 3/2loga(x²)? Because loga(x) + loga(x) = loga(x * x)?

[@monishrules6580]

No its 1/2logax

[@eamonburns9597]

@@monishrules6580 Ok, wouldn't it be: 1/2loga(x) + loga(x) = 3/2loga(x²)?

[@monishrules6580]

@@eamonburns9597 no lets say you get the 1/2 back onto x and then using the law add the logs you would get (x^1/2) × x which using laws of exponents we get x^1/2+1 which is x^3/2 and then you can just separate the 3/2 to get the same answer

[@eamonburns9597]

@@monishrules6580 Ok, then why does loga(x) + loga(x) = loga(x²)? I guess what I mean is: Why does x get squared if the log doesn't have a coefficient?

[@monishrules6580]

@@eamonburns9597 applying the same adding the log rules we get x × x and a thing multiplied by itself it just itself squared

[@agustindiaz3348]

8:05 could someone please enlighten me a little bit about the law he's using here? I didn't quite catch how the 1/x became -1 nor how the square root of x became 1/2. Thanks!

[@MarissaEYuret]

the square root of any number is the same as raising the number to 1/2

[@kurchak]

I took a different approach to Question #4: I went straight to the division rule by saying "log base a of (root x) / (1/x)" (Sorry, not sure how to type that in here. Then I multiplied by the reciprocal to get "log base a of (root x) * (x)" This was my final answer "log a (x * root x)" I put it in an equation solver and was surprised to see that this answer equates to the same thing as his "3/2 log a(x)". After staring at it for a while the intuition began to sit in, but if anyone else got my answer then just know that you are still correct. I only know this because I used symbolab to verify the answer.

[@zainislam554]

i didn't understand the last question! can anyone please help?

[@Liwet.]

Can three-halves of the log base a of x also be written as log a of the square root of x cubed?

[@rajabahmed6696]

sir do u take tutions? for class 11

[@vincentgiomontebon1199]

Your the best sir

[@quirkygirlboss]

You are a lifesaver

[@niketsrivastava2423]

Sir I have another solution to Q no.4 loga√x - loga1/x = loga(√x/1/x) = loga(√x^2) = logax

[@amanthegreat3257]

Same

[@chickenwingstick7594]

thank you 🙏🏻

[@saumyojitdas4212]

If log 2=0.30103,find the number of digits in 2^56.

[@balajisabbineni5610]

17

[@kyliedwyer6530]

im moving to this guys school

[@ohno8118]

I NEED HELP CUZ IM IN YEAR 3!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[@ayushsinha8553]

Last question could be solved more, i.e, 3/2log a x can be simplified to log a x√x.

[@nakitumizajashi4047]

As mr. Woo pointed out - it's always easier to multiply with plain number rather than calculating some weird power or multiplying with some weird number.

[@zainislam554]

i didn't understand the last question! can anyone please help?