Log Equation with Different Bases
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- Опубліковано 6 бер 2024
- In this lesson, I teach how to change the base of one log expression to match the other log express in order to combine them.
Also, the video show how to change 1 into a log expression.
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Nothing like a bit of Maths to start the day. Thanks Mr H!
I generally pick up a thing or two with your videos - with this one it was something that I should have known/realized years ago. You can square the base and square the thing you are taking the log of. Wow - why did I never use that? Thank you.
Weldone Sir 🔥🔥🔥
Thank you for your help and teaching 🌹
The nice explanation sir
Super keep well !!
Genius! Great teacher! ❤
One can use rule log base a of b + 1 = log base a (ab). It's quicker. But Mr. H's longer method is fantastic.
The goat
beautiful explanation
Good
👍
Sir but you can take square of -1/2 because log has coefficient of 2 in front and both the roots will satisfy the equation. Can you help me in resolving this doubt. Please
As it stand in the original problem the domain is x>2. When you put the 2 back to the power you are changing the domain. If it had started like (x-2)^2 to begin with then you would be right but that’s not how the original problem starts off. For example consider f(x)=x^2/x, most people will rush to simplify and say that means f(x)=x therefore the domain is all real numbers however they are wrong because x cannot equal 0 because it is not in the domain of f(x). And if you graph f(x) you will see it looks exactly like a line y=x, except x cannot equal 0. So we have a open dot at the point (0,0).
Jee question in shorts please 🙏
If one base is 7. another base is 3. How can I make them to be the same base?
Your clear, concise explanations, and selected problems have been quite useful in working with my tutoring students. Especially when I find bits missing from their school lessons, such this case, remembering to doublecheck the solutions in the original equation.
Thank you 📐🍀
I'm happy to hear that my videos are used to help your students.
1st 😊
isnt the 3/2 solution not possible, cause log9(3/2-2) gives an negative argument?
3:05
Why multiply times ¼? I don't see it.
To reduce and make computations of factoring easier
the greatest common factor of 8 and 36 is 4
Tedious for sure
(x-2)^2 X 9= x^2
... Good day, ... 2*LOG(b9)[ X - 2 ] + 1 = LOG(b3)[ X ] , where X > 2 ... 2*LOG(b3)[ X - 2 ] /( LOG(b3)[ 3^2 ] ) + LOG(b3)[ 3 ] = LOG(b3)[ X ] ... 2*LOG(b3)[ X - 2 ]/( 2*LOG(b3)[ 3 ] ) + LOG(b3)[ 3 ] = LOG(b3)[ X ] .... LOG(b3)[ X - 2 ] + LOG(b3)[ 3 ] = LOG(b3)[ X ] .... LOG(b3)[ 3*( X - 2) ] = LOG(b3)[ X ] .... 3*(X - 2) = X ... X = 3 .... checking in original equation ... S = { 3 } .... thank you for your presentation .... best regards, Jan-W
=> 2log9(x-2)-log3(x)=-1
2log9(x) - 4log9 -(log9(x)/log9(3)=-1
2log9(x) -4log9=-1+x/3=
2log9(x)/4log9=-1+x/3
x/2=-1+x/3 [(3x-2x)=-6]÷6
(1x=-6)/6 x/6=-6/6
x/6=-1 x=-6
All maths teachers are so dangerous 😡😡
The last answer
X=3/2 or X=3
Was like polynomials.
3/2 and 3 are the solutions
you could've just taken the square root of both sides once you got to (x-2)²•9=x2, since all of those numbers are perfect squares. you still get x=3 by doing this, and you don't have to worry about an extraneous solution.
Need to show why only 1 value of x is valid.
Can make it 9(x-2)^2 - x^2=0 then use diff of squares
(3[x-2]+x)(3[x-2]-x) =0
And get x=3/2 and x=3
Problem w just taking square root is you don’t know if the other solution is extraneous until you substitute it in.
Then there is me who used the change of base formula and did not get the extraneous solution.