Question

Binomial theorem

Answer 1

If the 6th term in the expansion of
\[\huge [\frac{ 1 }{ x ^{8/3}}+ x^2\log _{10}x]^8\] is 5600 , then find x

Answer 2

I have attempted this question , it may take some minutes to post

Answer 3

\(k\) th term of \((a+b)^n\) :

\[\binom{n}{\color{red}{k-1}} a^{n-(\color{Red}{k-1})} b^{\color{Red}{k-1}}\]

Answer 4

\(6\) th term of \((a+b)^8\) :

\[\binom{8}{\color{red}{6-1}} a^{8-(\color{Red}{6-1})} b^{\color{Red}{6-1}}\]

Answer 5

\[\huge \left(\begin{matrix}8 \\ 5\end{matrix}\right) x ^{\frac{ -512 }{ 27 }}.(x ^{2}\log _{10}x)^{5}\]=5600

Answer 6

\[\huge x ^{\frac{ -512 }{ 27 }}.(x ^{2}\log _{10}x)^{5} = 100\]

Answer 7

doesn't look correct

Answer 8

how did you get \(\large x^{\frac{-512}{27}}\) ?

Answer 9

yeah i felt i was making a terrible mistake somewhere

Answer 10

yes a very silly mistake actually

Answer 11

hmm,

x^{-8/3}^3

Answer 12

\[\large \left(a^m\right)^n = a^{m\times n}\]

Answer 13

-_- sry

Answer 14

\[\left(x^{-8/3}\right)^3 = x^{-8/3\times 3} = x^{-8}\]

Answer 15

\[\huge x ^{-8}.(x ^{2}\log _{10}x)^{5}=100\]

Answer 16

you should end up with
\[\large x ^{-8}\cdot (x ^{2}\log _{10}x)^{5} = 100\]

Answer 17

it is a equation now i can solve that , thank you

Answer 18

careful, it looks tricky...

Answer 19

\[\huge x^{2} . \log ^{5}_{10}x = 100\]

Answer 20

its a multiple choice , so by trial and error i got 10 hmm
how to do that subjectively

Answer 21

that looks good

Answer 22

x=10 makes the log thingy 1 and balances both sides so yeah..

Answer 23

so if it was a subjective paper how would we solve that

Answer 24

idk but it would be painful im sure

Answer 25

subjective paper?

Answer 26

yeah any method is not visible to me
not subjective paper

Answer 27

he wants to solve it analytically @perl

Answer 28

oh :)

Answer 29

but its good to know methods , my logarithm will also be revised

Answer 30

i think i got it

Answer 31

show me how

Answer 32

no it isn't working well

Answer 33

I will post it as a new question