For those of you telling sir that he could have solved the problem in three steps keep your mouth shut, he's a professor he knows wat he's doing and his intentions was to clearly illustrate for students that might not readily know how to go about solving this problem problem, thank you.
All humans have mistake except Lord Jesus and his mother virgn Mary and being professer doesn't make him knowing every thing and not to do mistake expect tirinty lord
I strongly disagree with you. In an exam students need the shortest possible mathematical way to solve problems. They don't need complicated methods and he being a professor should know that.
00:00 It is even more simple when you use the formula: log.(a^m) (x) = (1/m) * log.a(x) Hence you get: log.8(x) - log.16(x) = 0 log.(2^3)(x) - log.(2^4)(x) = 0 (1/3) * log.2(x) - (1/4) * log.2(x) = 0 Now it is quite simple: [ (1/3) - (1/4) ] * log(x) = 0 [ (4/12) - (3/12) ] * log(x) = 0 (1/12) * log(x) = 0 || *12 log(x) = 0 log.10(x) = 0 x = 10^0 x = 1 "1" is the solution, because it is in the Domain which is D: x>0
A man who makes totally time wasting short videos has 10 million subscribers in his channel but this channel is really awesome and has only approx 2 lakh. Not fair yaarrr
I love this. One thing I wanted to add to gues video is that being log8(x) and log16(x) have base 2 in common, we could rewrite it as log2^3(x) and log2^4(x) and identifying a log base value would be the denominator when the change of base is applied, we could then rewrite the equation as 1/3log2(x)-1/4log2(x) and then proceed from there.
Change of base, seems to suggest Gauge equation: unit conversion transform method. Conductance, energy, conversion and energy transform ascertained. Thanks and have a nice day, thank you Sam.
The log function is not defined on the interval ]-∞, 0]. You MUST state that BEFORE begining any calculus, hence excluding 0 as a solution. Thus checking the answer x=0 as you do is unnecessary (not to say wrong).
I love this. Two logs have different base and are equal. How's that? So that x = 1. Just because no other points where different-base logs can meet. Telling this basic story during nine minutes - great.
There is a short solution. You solve the right side y= log.16 (x) by taken it to the base (2*8) . y is only a placeholder for the watched term. So (2*8)^y=x eq(1). This can be simplified to 2^y*8^y=x (eq2). Now the left side is taken to base 8. So y=log.8 (x) becomes 8^y=x eq(3) , which can be written as 2^y*2^y*2^y=x. Now taken the 3. root from that: 2^y=3.root(x) (eq4) eq(3) and eq(4) we set in eq(2) and we get 3.root(x)*x=x. We can see the possible solution are x=0 and x=1. But x=0 is no solution, because the original equation get infinity. It remains x=1
These people were trying to outsmart the teacher.really? Why waste watching if you know already and know quick solution? ..this lecture is for those students or person who has difficulty and willing to understand it step by step. Just be humble guys, make ur own UA-cam channel..
Hi Sir, just wonder about this part x^4 = x^3, Couldn't you just divide both sides by x^3 so you could find just the true answer right away? Thanks anyway way.
@@war_reimon8343 You must factor it to find all the real solutions x^4 = x^3 x^4 - x^3 = 0 (x^3)(x - 1) = 0 (factoring x^3) The two solutions are: x^3 = 0 or x - 1 = 0 Therefore from the first equation x = 0 and from the second we get x = 1
Form the very begining you can take logx as a common factor then you will see that log x times 1/log8 minus 1/log16 equal zero which means that logx equal to zero therefore x equal 1
I have a Little doubt, sir At 6:30, you got X⁴ = X³, so as the bases are equal, we get 4 = 3 but 4 ≠ 3, I am a bit confused, could you please reply However, Thank you very much for this amazing video, I have been actively watching a lot of your videos and they have really amazed me! Keep it up! hope to see more!
If you try putting up the value of x as 1 or 0, the equation X³ = X⁴ will become valid, as we know, 1 to the power of n = 1; thus X³ = 1; X⁴ = 1; hence X³=X⁴ And. For x=0, 0^n = 0. Hence X³ =X⁴ =0 This won't work out on other values, except for 0 and 1... Hope you got my point....
Hello Eshwaraiah, here is the proof: Our LHS: log a basex=log b base x => use the change of base formula: (loga)/(logx) = (logb)/(logx) => both denominators get cancelled => loga = logb => use equality rule: a=b RHS. Hope it helps. If you want I can upload video for you as well. Take care and all the best😃
So nice of you Ramani dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
03:45 You should rather put this "(1/4)*log(x)" form the right side to the left side (with the sign "+" changed to "-"): (1/3)*log(x) = (1/4)*log(x) (1/3)*log(x) - (1/4)*log(x) = 0 [ (1/3) - (1/4) ] * log(x) = 0 [ (4/12) - (3/12) ] * log(x) = 0 (1/12) * log(x) = 0 || *12 log(x) = 0 log.10(x) = 0 x = 10^0 x = 1 "1" is the solution, because it is in the Domain which is D: x>0
It was obvious from the beginning that log x = 0, meaning x = 1, because the only way that raising x to two different powers gives an equality is if x = 1. This fellow used a sledgehammer to kill a fly. Far too complicated, tedious, and unnecessary.
I agree. And whatever the base is, log 0 is undefined, so the solution x=0 must be rejected before any calculus. (Always check the domain before starting !)
What i did at the end where we have: X^⅓ = X^¼ is: Divide both sides by X^¼ X^⅓/X^¼=1 Move the X^¼ to the numerator and get: X^⅓*X^-¼=1 X^1/12=1 (X^1/12)^12=1^12 X=1
I didn't do it that same way but still had an answer. I got to a part of 4INX - 3INX which is equal to zero. Then I subtracted 3 from 4, which which made it e of x is equal to zero. Then I had x to be one.
Dude! I did it the easier way out. First, what I did is, used the change of base formula ; then rewrote 16 as 2^4, and 8 as 2^3 ; then I used the power rule of logs "log(n^b) = b * log(n)" ; then I multiplied both of the fractions to get common denominators ; then I set just the numerator equal to zero after setting the whole fraction equal to 0 ; then I used the power rule of logs in reverse "b * log(n) = log(n^b)" ; then I used the quotient rule of logs ; then I used the quotient rule of exponents within the logs ; then I set log(x) to be equal to log(x) with base 10 ; then I converted to exponential form ; then noticed 10^0, which is equal to 1, and that is equal to x. What you are doing is the longer way out. you are checking for more possible solutions, but they are extraneous, so it would have been better if you did what I did no?
I think we can solve this problem in another way avoiding all the complicated calculations Logically how can the power of 8 and 16 be same for the same value unless the value is 1 and power is 0 hence x has to be equal to 1
Thank you so much luyando mweemba for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep supporting my channel. Please stay connected. Take care dear 😃
God loves you and he wants to save everyone, but in order for him to do that, you need to repent and be baptized. Also share his gospel with everyone you come in to contact with and keep his commandments 🙏🏾😘
![Logarithm Change of Base Formula & Solving Log Equations - Part 1 - [7]](http://i.ytimg.com/vi/BIgf3pH9PTY/mqdefault.jpg)
For those of you telling sir that he could have solved the problem in three steps keep your mouth shut, he's a professor he knows wat he's doing and his intentions was to clearly illustrate for students that might not readily know how to go about solving this problem problem, thank you.
Thank you so much! Please keep supporting my channel. Kind regards 😀
All humans have mistake except Lord Jesus and his mother virgn Mary and being professer doesn't make him knowing every thing and not to do mistake expect tirinty lord
I strongly disagree with you. In an exam students need the shortest possible mathematical way to solve problems. They don't need complicated methods and he being a professor should know that.
Just because he's a professor doesn't make everyone else an idiot.
00:00 It is even more simple when you use the formula:
log.(a^m) (x) = (1/m) * log.a(x)
Hence you get:
log.8(x) - log.16(x) = 0
log.(2^3)(x) - log.(2^4)(x) = 0
(1/3) * log.2(x) - (1/4) * log.2(x) = 0
Now it is quite simple:
[ (1/3) - (1/4) ] * log(x) = 0
[ (4/12) - (3/12) ] * log(x) = 0
(1/12) * log(x) = 0 || *12
log(x) = 0
log.10(x) = 0
x = 10^0
x = 1
"1" is the solution, because it is in the Domain which is D: x>0
I came up with the same solution as yours, before watching the video and comments. It's quicker and without unuseful steps.
Excellent tutorial covering, ahem, a lot of bases!
No pun was intended!
A man who makes totally time wasting short videos has 10 million subscribers in his channel but this channel is really awesome and has only approx 2 lakh. Not fair yaarrr
I love this. One thing I wanted to add to gues video is that being log8(x) and log16(x) have base 2 in common, we could rewrite it as log2^3(x) and log2^4(x) and identifying a log base value would be the denominator when the change of base is applied, we could then rewrite the equation as 1/3log2(x)-1/4log2(x) and then proceed from there.
0pplp0pd0lo 0poppp0pp0
And further take value of log2(x) to be t solve t it only gives one value which is 1 and solve less lengthy
Change of base, seems to suggest Gauge equation: unit conversion transform method. Conductance, energy, conversion and energy transform ascertained. Thanks and have a nice day, thank you Sam.
The log function is not defined on the interval ]-∞, 0]. You MUST state that BEFORE begining any calculus, hence excluding 0 as a solution. Thus checking the answer x=0 as you do is unnecessary (not to say wrong).
I love this. Two logs have different base and are equal. How's that? So that x = 1. Just because no other points where different-base logs can meet. Telling this basic story during nine minutes - great.
i liked and watched the video sir
Thanks. You are awesome.
There is a short solution.
You solve the right side y= log.16 (x) by taken it to the base (2*8) . y is only a placeholder for the watched term.
So (2*8)^y=x eq(1). This can be simplified to 2^y*8^y=x (eq2).
Now the left side is taken to base 8. So y=log.8 (x) becomes 8^y=x eq(3) , which can be written as 2^y*2^y*2^y=x.
Now taken the 3. root from that: 2^y=3.root(x) (eq4)
eq(3) and eq(4) we set in eq(2) and we get 3.root(x)*x=x.
We can see the possible solution are x=0 and x=1. But x=0 is no solution, because the original equation get infinity.
It remains
x=1
Really, exciting Vedio 🔥😇
These people were trying to outsmart the teacher.really? Why waste watching if you know already and know quick solution? ..this lecture is for those students or person who has difficulty and willing to understand it step by step. Just be humble guys, make ur own UA-cam channel..
This guy is an excellent teacher.
Hi Sir, just wonder about this part x^4 = x^3, Couldn't you just divide both sides by x^3 so you could find just the true answer right away? Thanks anyway way.
x^4=x^3 is not possible
@@albertmendoza1468 It's possible if x is equal to 0 or 1
When x=0,you are dividing by 0.
@@tigerbeast3406 0 is not possible because it is undetermined. So it leaves only possible the one.
@@war_reimon8343 You must factor it to find all the real solutions
x^4 = x^3
x^4 - x^3 = 0
(x^3)(x - 1) = 0 (factoring x^3)
The two solutions are:
x^3 = 0 or x - 1 = 0
Therefore from the first equation x = 0 and from the second we get x = 1
Sir, you don't need to do that long since log_n(1)=0
I used log.2 when using the change of base formula, this eliminated many of the steps. i only arrived at 1 solution, x = 1.
May God almighty bless you
Sir, usually when doing this problem you also have to add two with log of x
At starting point can be eliminate logx from boths side, then how can get x value?
At 1:34, you have Log X on both sides. If we divide by this we will have 1/Log 8 = 1/Log 16.
How can this be?
@@XJWill1 Thanks!
I am Bangladeshi 🥰🥰🥰 thanks a lot ❤️
I need some of your previous videos on logs
u have made it an equation history.it could be solved more easierly
At 2:59 log (x)/a = log(x)/b is impossible unless log (x) = 0 or x=1.
Sir the solution is maximum 3 steps only ....
We used to get points taken off for "excessive math!"
hes trying his best, so dont be so rude
can you explain to me how it is easier? I am in a math class at my college and I want to understand how it can be done in 3 steps
Form the very begining you can take logx as a common factor then you will see that log x times 1/log8 minus 1/log16 equal zero which means that logx equal to zero therefore x equal 1
@@jennybenson9562 o .i see.u r in college but also u r not able to use a simple logarithmic equation
The lesson is well understood sir,thanks a lot.
x=1
I watched and liked it
Thanks. You are awesome.
So far so good 👍
Thank you so much .
I have a Little doubt, sir
At 6:30, you got X⁴ = X³, so as the bases are equal, we get 4 = 3 but 4 ≠ 3, I am a bit confused, could you please reply
However, Thank you very much for this amazing video, I have been actively watching a lot of your videos and they have really amazed me! Keep it up! hope to see more!
If you try putting up the value of x as 1 or 0, the equation X³ = X⁴ will become valid, as we know, 1 to the power of n = 1; thus X³ = 1; X⁴ = 1; hence X³=X⁴
And. For x=0, 0^n = 0. Hence X³ =X⁴ =0
This won't work out on other values, except for 0 and 1...
Hope you got my point....
Sir please give solution for this
If log a basex=log b base x thena=b if where a>0^b>0^ x>0 x is not equal to 1
Hello Eshwaraiah, here is the proof:
Our LHS: log a basex=log b base x => use the change of base formula: (loga)/(logx) = (logb)/(logx) => both denominators get cancelled => loga = logb => use equality rule: a=b RHS.
Hope it helps. If you want I can upload video for you as well. Take care and all the best😃
Lo scopo dell'esercizio proposto e la sua soluzione, è anche di mostrare molte belle proprietà dei logaritmi e le molte ed eleganti vie del calcolo.
Thanks a lot sir,
Fantastic explanation 👍
So nice of you Ramani dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Can you say that cause X^3 = X^4 so it have to be 1 and 0 otherwise its not possible?
Instead of using base 10 it is easier if you use base 2 on the change of base
X should only be equal to 1 sir as log of zero to any base is under define
Yep, and it can only be one value as it's only x^1
03:45 You should rather put this "(1/4)*log(x)" form the right side to the left side (with the sign "+" changed to "-"):
(1/3)*log(x) = (1/4)*log(x)
(1/3)*log(x) - (1/4)*log(x) = 0
[ (1/3) - (1/4) ] * log(x) = 0
[ (4/12) - (3/12) ] * log(x) = 0
(1/12) * log(x) = 0 || *12
log(x) = 0
log.10(x) = 0
x = 10^0
x = 1
"1" is the solution, because it is in the Domain which is D: x>0
It was obvious from the beginning that log x = 0, meaning x = 1, because the only way that raising x to two different powers gives an equality is if x = 1. This fellow used a sledgehammer to kill a fly. Far too complicated, tedious, and unnecessary.
Yes. The charm of mathematics is in finding the shortest path!
I agree. And whatever the base is, log 0 is undefined, so the solution x=0 must be rejected before any calculus. (Always check the domain before starting !)
@@geoellinas I think the charm of mathematics is paths
@@john-phimcmelley6422 For mathematicians yes. But for the students I believe what I said.
It amazes me how many answers are -1, 0, or 1.
Sir, can you do this problem
What i did at the end where we have:
X^⅓ = X^¼ is:
Divide both sides by X^¼
X^⅓/X^¼=1
Move the X^¼ to the numerator and get:
X^⅓*X^-¼=1
X^1/12=1
(X^1/12)^12=1^12
X=1
Thanks for that
Had a confusion about the base changing of a logarithm. It totally solved the problem. Appreciate it👏
I didn't do it that same way but still had an answer. I got to a part of 4INX - 3INX which is equal to zero. Then I subtracted 3 from 4, which which made it e of x is equal to zero. Then I had x to be one.
Only 1 fulfills the equation.
Why do you go long distance
Dude! I did it the easier way out. First, what I did is, used the change of base formula ; then rewrote 16 as 2^4, and 8 as 2^3 ; then I used the power rule of logs "log(n^b) = b * log(n)" ; then I multiplied both of the fractions to get common denominators ; then I set just the numerator equal to zero after setting the whole fraction equal to 0 ; then I used the power rule of logs in reverse "b * log(n) = log(n^b)" ; then I used the quotient rule of logs ; then I used the quotient rule of exponents within the logs ; then I set log(x) to be equal to log(x) with base 10 ; then I converted to exponential form ; then noticed 10^0, which is equal to 1, and that is equal to x.
What you are doing is the longer way out. you are checking for more possible solutions, but they are extraneous, so it would have been better if you did what I did no?
Mr sanel he is a professor of what. I can also call you professor
I think we can solve this problem in another way avoiding all the complicated calculations
Logically how can the power of 8 and 16 be same for the same value unless the value is 1 and power is 0 hence x has to be equal to 1
thanks
You're welcome Tim!
Thank you. You are awesome😀
Keep smiling😊 Enjoy every moment of your life 🌻
logx b2/3-logxb2/4=0=>logxb2=0=>x=2^0=1ans
Good
I think x = {0,1,2,3,4,5,6,7} , what is your idea
Why the answer not x=1?
(log x)/(log 8) = (log x)/log16
Log 8 = log 16
8 = 16
Where did I make a mistake professor
I am understanding
Log of x base2=0=>2^0=x=>x=1
3^(x-1) =5
Solution:
(1) log8(x) - log16(x) = 0
According to logc(y) = loga(y)/loga(c) = same base (number/base), the following applies:
log16(x) = log8(x)/log8(16) | applied in equation (1), results in:
(1a) log8(x)-log8(x)/log8(16) = 0 |*log8(16)≠0 ⟹
(1b) log8(x)*log8(16)-log8(x) = 0 ⟹
(1c) log8(x)*[log8(16)-1] = 0 |/[log8(16)-1]≠0 ⟹
(1d) log8(x) = 0 ⟹ x = 8^0 = 1
Lösung:
(1) log8(x)-log16(x) = 0
Nach logc(y) = loga(y)/loga(c) = gleiche Basis (Numerus/Basis) gilt:
log16(x) = log8(x)/log8(16) | in Gleichung (1) angesetzt, ergibt:
(1a) log8(x)-log8(x)/log8(16) = 0 |*log8(16)≠0 ⟹
(1b) log8(x)*log8(16)-log8(x) = 0 ⟹
(1c) log8(x)*[log8(16)-1] = 0 |/[log8(16)-1]≠0 ⟹
(1d) log8(x) = 0 ⟹ x = 8^0 = 1
Amazing..
Thank you so much luyando mweemba for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep supporting my channel. Please stay connected. Take care dear 😃
log x/3log 2 =log x/4log 2, so log x=0, x=1😃
I almost solved this but got stuck at the end cause of splitting of x4-x3😅
God loves you and he wants to save everyone, but in order for him to do that, you need to repent and be baptized. Also share his gospel with everyone you come in to contact with and keep his commandments 🙏🏾😘
I got answer as 12
Where will be 12
C'est une farce?
Ouais c’est farce, x=1
X is 1
log(🐭) ×log(🐭) =[log(🐭)] ×2 or [log(🐭)]?😔😔
Can somebody solve 4^ = 2^x
X=2
logx/log8=logx/log16
logx=0
x=1
It was obvious
Why complicate things
Elementary school math
Ok
حلك غير صحيح ﻻن المعادلة ﻻتتحقي اﻻ عندماx تساوي الصفر
diwiiided buy
forse 1
solo a vederla mi sembra non abbia soluzioni
mind twisted already
it long
Why is it so confusing😂
Pliz h
Воды много, автор объём нагоняет...
Too many steps.
Wrong dull teacher
haaa
log x 8 = 3/2 how to solve
x^(3/2)=8
x=8^(2/3)
8^(2/3)=(8^(1/3))^2=2^2=4
x = 4
and, 4^(3/2)=8
Q.E.D
(actually, this problem's solution is logarithme's definition.)
@@역학적에너지-i9z xnxc
x=1
Can somebody solve 4^x = 2^x
(2^2)^x = 2^x
2^2x = 2^x
2x = x
2x - x = x - x
x = 0