# Thread: Binomial expansion?

1. ## Binomial expansion?

At what $\displaystyle x$ the $\displaystyle 6^{th}$ term in the expansion of the binomial:
$\displaystyle \left[\sqrt{\left(2^{\log (10 - 3^x)}\right)} + \sqrt [5]{\left(2^{(x - 2)\log 3}\right)}\right]$
is equal to 21 if it is known that the binomial coefficients of the $\displaystyle 2^{nd},3^{rd}$ and the $\displaystyle 4^{th}$ terms in the expansion represent respectively the first, third and fifth terms of an arithmetic progression?

(A) $\displaystyle x = 6$
(B) $\displaystyle x = 2$
(C) $\displaystyle x = -2$
(D) $\displaystyle x = 11$

2. Hello,

Maybe I'm mistaking... but if x=6, then $\displaystyle 10-3^6<0 \Rightarrow \log(10-3^6)$ is not defined.
Same thing for x=11.

------------------------
If x=2
then $\displaystyle \log(10-3^x)=\log(1) \Rightarrow 2^{\log(10-3^x)}=1$

And $\displaystyle 2^{(x-2)\log(3)}=1$

So the expression is 2.
------------------------
Therefore, x=-2 ??

3. Originally Posted by Moo
Hello,

Maybe I'm mistaking... but if x=6, then $\displaystyle 10-3^6<0 \Rightarrow \log(10-3^6)$ is not defined.
Same thing for x=11.

------------------------
If x=2
then $\displaystyle \log(10-3^x)=\log(1) \Rightarrow 2^{\log(10-3^x)}=1$

And $\displaystyle 2^{(x-2)\log(3)}=1$

So the expression is 2.
------------------------
Therefore, x=-2 ??
Ahh! Back to basics Nice.