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.
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).
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 📐🍀
If you did a root before you opened the exponent and you do a root on the whole expression you would still get x = 3 but still lose the answer of 2/3 🧐 you would lose answers because it is better to pay attention before you open a root and get. To see that I don't lose answers for some reason I never lost an answer 😏
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.
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.
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.
Thank you for your help and teaching 🌹
Weldone Sir 🔥🔥🔥
Genius! Great teacher! ❤
The nice explanation sir
Super keep well !!
Good
beautiful explanation
The goat
👍
1st 😊
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.
isnt the 3/2 solution not possible, cause log9(3/2-2) gives an negative argument?
=> 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
Tedious for sure
... 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
All maths teachers are so dangerous 😡😡
(x-2)^2 X 9= x^2
Why not take the square root of both sides of the equation (x-2)^2.9=x^2. That gives the only solution of x=3
If you did a root before you opened the exponent and you do a root on the whole expression you would still get x = 3 but still lose the answer of 2/3 🧐 you would lose answers because it is better to pay attention before you open a root and get. To see that I don't lose answers for some reason I never lost an answer 😏
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.