Solving log equations with different bases can be tricky. In this video, one can learn how to change the base of one log so that it matches the base of the other log expression.
I played around with these rule myself, plugging in different numbers and possibilities. It always works! I wonder why we didn’t learn this particular rule of logs back in the 90s when I was in highschool and early college?
@@TheMathManProfundities Thanks. Let me replicate my previous comment: You know, the problem with these sorts of problems is that there are a ton of little rules for logs that can be used, and one could memorize them all and when to use them. But that's what makes "math" a drag; the alternative is to know a small set of rules that others can be derived from. Like in this problem, why not just use the change of base rule, and forget the special base to a power rule?
You know, the problem with these sorts of problems is that there are a ton of little rules for logs that can be used, and one could memorize them all and when to use them. But that's what makes "math" a drag; the alternative is to know a small set of rules that others can be derived from. Like in this problem, why not just use the change of base rule, and forget the special base to a power rule?
@@mrhtutoring Yes - you just use the change of base on all terms on the left and use base 2, as 逸園-無毒果園 comments below (above ?). Don't get me wrong - I enjoy your videos - thanks.
You can't just say that because the powers are the same, x = 16. For example, x²=2² has two solutions, x = ±2. In this case there are actually seven solutions, the other 6 being complex numbers. So x = 16e^(2kπi/7) ∀k∈ℤ∩[0,6].
We can also do it in this way from firstt log x to the base 16, just write it as 4^2 and then by the property we can write it as half of log x to the base of 4. Same way do it again and u get base 2. Same with the secons log term and then turn that co efficients to the powers of x and i guess its now solvable. Correct me if im mistaken somewhere because i just learned log properties and its basics. I would love to see someone correct me.
Did not know this property of logs. Makes perfect sense, given some thought. Thank you!
Yeah you did - it's just an application of the change of base rule and argument to a power rule.
I played around with these rule myself, plugging in different numbers and possibilities. It always works! I wonder why we didn’t learn this particular rule of logs back in the 90s when I was in highschool and early college?
This is such a save when you have a math test on log tomorrow 😭❤️🌟
Thank you so much!
I will teach this rule next year to my students. Thanks.
All the best~
Hail from Brazil! Nice solution
"Did not know this property of logs": me neither. Wow!
Wow. I'm always learning new logarithm properties. I wonder how many logarithm properties exist in math. Is it infinite? 🤔
They are NOT infinite but we've quite a bunch of them. According to my experience logaithm properties are better studied by practising.
This is just a result of other laws i.e. logₐx=ln x/ln a=bln x/(bln a)=ln xᵇ/ln aᵇ=log_{aᵇ} (xᵇ).
@@TheMathManProfundities Thanks. Let me replicate my previous comment: You know, the problem with these sorts of problems is that there are a ton of little rules for logs that can be used, and one could memorize them all and when to use them. But that's what makes "math" a drag; the alternative is to know a small set of rules that others can be derived from. Like in this problem, why not just use the change of base rule, and forget the special base to a power rule?
@@jonahansen My point exactly, the only log rules you really need to know are log(aᵇ)=b log(a) and log(AB)=log(A) + log(B).
That was impressive.
Very good thanks for sharing
Excellent method
Excellent! Thank you!
log16(x)+log4(x)+log2(x)=7
log2(4x)+log2(2x)+log2(1x)=7
log2(4•2,52)+log2(2•2,52)+log2(1•2,52)
log2(10,08)+log2(5,04)+log2(2,52)=
3,33'+2,33'+1,33'=7
x=2,52
Thank you for your teaching 🌹
I like your teaching 🌹
You are very kind teacher ❤
It was awesome.
can you teach me on about integretion and limitations.
This is brilliant.
Thanks teacher
You know, the problem with these sorts of problems is that there are a ton of little rules for logs that can be used, and one could memorize them all and when to use them. But that's what makes "math" a drag; the alternative is to know a small set of rules that others can be derived from. Like in this problem, why not just use the change of base rule, and forget the special base to a power rule?
If you use the change of base formula, can you solve for x without a calculator?
@@mrhtutoring Yes - you just use the change of base on all terms on the left and use base 2, as 逸園-無毒果園 comments below (above ?). Don't get me wrong - I enjoy your videos - thanks.
Thanks
base=2, we have (1/4)log2(x)+(1/2)log2(x)+log2(x)=7,so log2(x)=4, x=2^4=16
That's a good one. Thank you.
Great
Nice
Am from Uganda en i love the way you calculate math problems. Am requesting u to use us differential equations thank u very much ❤❤❤❤❤❤❤❤❤❤0:00
thank you sir 🙏
Genius guy
You can't just say that because the powers are the same, x = 16. For example, x²=2² has two solutions, x = ±2. In this case there are actually seven solutions, the other 6 being complex numbers. So x = 16e^(2kπi/7) ∀k∈ℤ∩[0,6].
Sir, please upload some videos on integration.
We can also do it in this way from firstt log x to the base 16, just write it as 4^2 and then by the property we can write it as half of log x to the base of 4. Same way do it again and u get base 2. Same with the secons log term and then turn that co efficients to the powers of x and i guess its now solvable. Correct me if im mistaken somewhere because i just learned log properties and its basics. I would love to see someone correct me.
interesting, I didnt know this was a property of logs. very useful though, thank you.
You're welcome!
Very good
Thanks
log16(x) + log4(x) + log2(x) = 7
log2⁴(x) + log2²(x) + log2(x) = 7
1/4 (log2(x)) + 1/2 (log2(x)) + log2(x) = 7
log2(x)¼ + log2(x)½ + log2(x) = 7
log2(x¼.x½.x) = 7
(x¼)⁷ = 2⁷
x¼ = 2
x = 2⁴
x = 16
Nice
How come
log2(x+x2+x4)
@@omargameover5438
?
@@ayunda.alicia
By Using properties of logarithm
Our math teacher never taught us such details of logs
@@sajidrafique375
Yeah you are right although where are you from ?
@@AbcdAbcd-p5e Santa Barbara city college , caifornia but i was a foreign student there from pakistan ..
🤗
Maybe at the end just 7logx = 7
Could he have used the base transfer formula instead?
How many people accidentally square or power to 4 the 7 on the other side.
Very helpful.
If only you don't use writing in colours other than white and yellow. Both colours are perfect and visible
Ok next time
he x should be 2 also if u just write power before log and put it on x head it be 2 powe 7 is x pow3er 7 so x is 7
Sure….it’s easy for a math genius…..