@@wes9627 Recall that Jethro attended Oxford. And he was able to do advanced cipherin' whilst in the 5th grade... a class which he repeated for many consecutive school years until he towered over all his classmates and teachers.
This can be difficult for anyone who has not solved any problems like this. On an exam with limited time, practice solving simple problems like this may be the difference between passing or failing.
@@wes9627 In reality, either we conceive the University as a place of thought (and then we attach ourselves to ideas) or as a rodeo where apparent performance takes the place of selection system. My dear France, thanks to "the stupidity of its elites", shows today an example not to follow. Frédéric CHOPIN has proved that the best way to play badly on the piano is to recite scales and arpeggios at full speed... He must have read Mark TWAIN... Do you agree?
The more we detail unnecessary calculations, the more confused we are !!! If you want your listeners to ASSIMILATE and USE correctly the rules of exponents, start by stating them verbally ! In order to drive the nail right before starting, illustrate each principle by 3 or 4 simple examples. a) a multiplication is expressed as an addition ; b) the putting in a radical n-th is expressed by the division by the index n of this radical ; c) the structure of the proposed expression is precisely a monome of 5. Thus, by compiling the exponents as and when, from right to left, we obtain WITHOUT ANY INTERMEDIARY the sum : [ 3x / 2 / 2 / 2 ] + [ 2x / 2 / 2 ] + [ x / 2 ] = 3x/8 + 2x/4 + x/2 = 11x/8 = 3 ==> x = 24/11
Good show. I always try to solve a problem first and then compare my method/results with the given solution. In this case, in my head I got x=24/11, but had difficulty following the solution to compare my method/results. In my head 3→3/2→7/2→7/4→11/4→11/8→x=3/(11/8)=3(8/11)=24/11. I'm in my 80s and solve problems like this to stave off dementia, definitely not to get into any university.
@@wes9627 Impeccable. Only the elderly know the cost of lost time... In order to stave off your dementia, definitely NOT to show off as French Master Renart, can you proove (3 lines ?) that the prime number 641 divides 2^32 - 1 ? It's tea time...
LHS
=5^{x/2+2x/4+3x/8)
= 5^{(4x+4x+3x)/8}
=5^(11x/8)=RHS
=125=5^3; =>11x/8=3;
=> x = 24/11
Relatively trivial. Can I ask what problem you came across requires solving this particular equation?
Instead of counting sheep at 3am I do my goes-in-tas or ciphering as Jethro Bodine would say. No need for a reason.
@@wes9627 Recall that Jethro attended Oxford.
And he was able to do advanced cipherin' whilst in the 5th grade... a class which he repeated for many consecutive school years until he towered over all his classmates and teachers.
125^x=(5^3)^x=
(5^x)^3
25=(5^x)^2
Đặt y=5^x
this is a clickbait, thit is not hard at all
This can be difficult for anyone who has not solved any problems like this. On an exam with limited time, practice solving simple problems like this may be the difference between passing or failing.
@@wes9627 In reality, either we conceive the University as a place of thought (and then we attach ourselves to ideas) or as a rodeo where apparent performance takes the place of selection system. My dear France, thanks to "the stupidity of its elites", shows today an example not to follow. Frédéric CHOPIN has proved that the best way to play badly on the piano is to recite scales and arpeggios at full speed... He must have read Mark TWAIN... Do you agree?
It came out to 24/11, but when you plug this vack into the original expression, it is diffucult to derive the sought-after 125 !!
The more we detail unnecessary calculations, the more confused we are !!! If you want your listeners to ASSIMILATE and USE correctly the rules of exponents, start by stating them verbally ! In order to drive the nail right before starting, illustrate each principle by 3 or 4 simple examples.
a) a multiplication is expressed as an addition ;
b) the putting in a radical n-th is expressed by the division by the index n of this radical ;
c) the structure of the proposed expression is precisely a monome of 5.
Thus, by compiling the exponents as and when, from right to left, we obtain WITHOUT ANY INTERMEDIARY the sum :
[ 3x / 2 / 2 / 2 ] + [ 2x / 2 / 2 ] + [ x / 2 ] = 3x/8 + 2x/4 + x/2 = 11x/8 = 3 ==> x = 24/11
Good show. I always try to solve a problem first and then compare my method/results with the given solution. In this case, in my head I got x=24/11, but had difficulty following the solution to compare my method/results. In my head 3→3/2→7/2→7/4→11/4→11/8→x=3/(11/8)=3(8/11)=24/11. I'm in my 80s and solve problems like this to stave off dementia, definitely not to get into any university.
@@wes9627 Impeccable. Only the elderly know the cost of lost time...
In order to stave off your dementia, definitely NOT to show off as French Master Renart, can you proove (3 lines ?) that the prime number 641 divides 2^32 - 1 ? It's tea time...
let u=5^x , u^(3)^(1/2)=u^(3/2) , u^(3/2)*u^(4/2)=u^(7/2) ,
u^(7/2)^(1/2)=u^(7/4) , u^(7/4)*u^(4/4)=u^(11/4) , u^(11/4)^(1/2)=u^(11/8) , u^(11/8)=5^3 , /()^(8/11) , u=(5^3)^(8/11) , u=5^(24/11) ,
5^x=5^(24/11) , result , x=24/11 ,