How to Solve Advanced Logarithmic Equations Involving Square Roots and Cube Roots: Easy Explanation
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- Опубліковано 14 січ 2025
- Learn how to Solve Challenging Logarithmic Equations that involve cube roots, square roots, and more than One Radical. Also, check your answer for extraneous solutions. Step-by-Step Tutorial by PreMath.com
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Good teaching pedagogy, u didn’t miss steps that usually other teachers presume students knows it
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Wow! I went back 4 years and the teaching is still very detailed and thorough
Great teacher, I admire your work
Very good explanation without jumping the steps. I say this because, people who know to work out , simply jump the steps.
Thanks for the input.
So nice of you Ramani! You are awesome 👍 I'm glad you liked it! Take care dear and stay blessed😃
When you square both sides, you need to make sure both sides have the same sign. The right side is a sqrt so it's never negative, therefore it's important to remember the domain for our calculations is u/3>=0. In this case it didn't make a difference, but it should be noted and considered
great explanation. though I square both sides to get log x^2/3 = log x, . and glad to see it has two solutions.
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Thank you so much, really appreciated this video
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Have a very happy and blessed New Year!
Thank you so much! Awesome video, very helpful
i enjoyed this one. Thanks for your time.. Wish you an early Happy Holidays...😀
Excellent
Well explained
#DefinationsForStudents
Extended beautifully
thank you sir for the good explanation!
Great exercise, using substituion.
thank you
What I can't understand here is why on LHS we move cube root to front log while on RHS we don't move sq root.
Pls help
It is because of what was being raised to a power. On the LHS it was only the variable X that was raised to the 1/3 power. On the RHS it was the whole expression logx that was being raised. The rule for bringing the exponent in front of the log only works when it is the variable being raised to a power, not the whole expression.
On LHS, the "cuberoot" operator is inside the log.
On RHS, the "square root" operator is already outside the log.
I did it in my mind in few seconds
Excellent
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Sir, i think (log x)^1/2 doesn't equal to log x^1/2
Tks teacher.
Pls sir why RHS log is concentrated in the parenthesis while LHS not.
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@@PreMath
Kindly if you can answer my question
Tks for yr good cooperation
Hi guys pls I need your help
Why exponent of the LHS was moved to the front logx while RHS exponent was not moved to the front logx
@@glasssmirror2314 I think it was already moved to the front from the beginning.
Since the RHS is √logx you can rewrite it as (logx)^1/2. Normally the root only covers the 'x'. For instance, log√x which also where you can use the properties of logs, that is to say, log(x)^1/2, move it to the front as 1/2logx or 1logx/2. So the difference was only where the exponent is situated at from the given, that RHS is simply (logx)^1/2 rather than log(x)^1/2
7:04 Yes I understand it is a huge number. But it is a power of 10! It is very easily evaluatable! It is 1 billion. Which is "1000000000". Notice there are 9 zeros there? That is the exponent. That is all, the way to evaluate a power of 10! It is a huge number, but easy to evaluate.
Thats not the point lmao everyone knows what a billion is
That's great
Vvnice true thanks
But x cant be equqls to 1 bcz logs rules
Log can't have base 1, but the number in log can be any positive number
Be careful when the power root is the even order, the answer is can be positive number and negative number. Such as sqrt(9) = +3 and -3.
We typically don’t say that a definite value, such as sqrt(16), is equal to plus or minus 4. This is because we generally want to avoid having multiple solutions in one definite notation. But when you want to solve x^2 = 16, then you can say that x is equal to plus or minus 4 as when we solve equations we want to find all the possible values of x since its value isn’t known. And on certain occasions such as this (x-2)(x+2) = 0, there are two *possible* values that x can have.
Great
👏👏👏👏👏👏
Easy question
3√(x) = x^(1/3), so log(3√x) = (log(x))/3. Let y = log(x). We have y/3 = √y, so y=9, and x=10^9=1,000,000,000. Took about one minute, not 12.
First of all, you missed the second solution. X=1 is also a solution, even though it looks trivial. You should never cancel any variable on both sides unless you're certain they're not equal to zero, which in this case, they were, as log1 to any allowed base is 0. Moreover, you think that guy can't solve this fast? This is an easy question, but he's teaching. Not everyone knows how to do this stuff. Could I have solved it quicker? Definitely. But he needs to make sure everyone follows, not just people who already know how to handle logarithmic functions.
He was in teaching mode, demonstrating log rules as he went along, and then proved both answers. It was very nicely done and didactic. Your criticism is absurd.
As both Singh and Amram said, he was teaching so he has to assume that everyone is not at the same level.
And as Einstein said 'If you can't explain it to a six-year-old, you don't understand it yourself '. In other words, sometimes you will need to be able to explain it, in so many ways, that you are able to explain it, even to, those who might not pick it up, as easy as others or is at the same level when it comes to logarithm. And that what he is attempting to do.
#log #logarithm
Tòoooooooo goooooood
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1; 10⁹
#cuberoot
log base 10 (x)
Cube root of x = x^(1÷3)
#substitution
u = log (x)
log (x)
X=10^9 and x=1
x variable
First: what Log do you mean ? Then(!) ask what x is a solution ?!
u^(2) = 9u
x = 10^(9)
u^(2) - 9u = 0
log (1000) = 3
Square root of 9 = 3
And no any advance log
x = 1
log (1) = 0
Simple problem not advance
u = 9
u = 0
Sir its 3 steps only
Dear Math S, you are absolutely correct. There are many different ways to solve any given problem. Thanks for the feedback. Kind regards :)
Wothless
3=9root
No
It s - 3'3