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.
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?
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.
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].
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.
Didn't know this, too. You teach with such clarity.
I will teach this rule next year to my students. Thanks.
All the best~
Hail from Brazil! Nice solution
Excellent! Thank you!
This is such a save when you have a math test on log tomorrow 😭❤️🌟
Thank you so much!
It was awesome.
can you teach me on about integretion and limitations.
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
Thanks teacher
That's a good one. Thank you.
That was impressive.
This is brilliant.
Thank you for your teaching 🌹
I like your teaching 🌹
You are very kind teacher ❤
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?
Very good thanks for sharing
Excellent method
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).
thank you sir 🙏
Nice
Great
Genius guy
"Did not know this property of logs": me neither. Wow!
Sir, please upload some videos on integration.
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
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.
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
Very good
Thanks
logxbase4 can be written as lnx/ln4 i.e lnx/(2ln2) ie 1/2 logxbase2
so above equation becomes
1/4logxbase2 + 1/2 logxbase2 + logxbase2 i.e 7/4logxbase2 = 7 ie 1/4logxbase2 = 1 ie logxbase2 = 4 ie x = 16
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
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].
interesting, I didnt know this was a property of logs. very useful though, thank you.
You're welcome!
🤗
How many people accidentally square or power to 4 the 7 on the other side.
Could he have used the base transfer formula instead?
Maybe at the end just 7logx = 7
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 ..
log16_x+log4_x+log2_x
=log16_x+2log16_x+4log16_
=7log16_x=7
log16_x=1
x=16
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…..
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