The binomial coefficients of the expansion (a+b)^m are the consecutive tems of an arithmetical sequence.The term of expansion that is 21 is having...

We'll start using the condition from enunciation that the
binomial coefficients of the given expansion are in arithmetical
progression.

Since it is not indicated what are the terms,
we'll suppose that they are the 2nd, 3rd and the 4th
terms.

C(m,2) = [C(m,1) +
C(m,3)]/2

We'll re-write the condition
above:

m(m-1) = m +
m(m-1)(m-2)/6

We'll multiply by 6 all the
terms:

6m(m-1) = 6m +
m(m-1)(m-2)

6m(m-1) = m[6 +
(m-1)(m-2)]

We'll divide by
m:

6m - 6 = m^2 - 3m + 8

We'll
move all terms to one side and we'll combine them:

m^2 - 9m
+ 14 = 0

We'll apply the quadratic
formula:

m1 = (9+sqrt25)/2

m1
= (9+5)/2

m1 = 7

m2 =
(9-5)/2

m2 = 2

But it is
impossible for the value of m to be smaller than the value of k, that is 5, from the
condition from enunciation, that one term is containing
b^5.

Putting m = 7, we'll write the formula of the genertal
term of the expansion:

T6 =
C(7,5)*a62*b^5

21 =
7(7-1)/2*2^lg(10-3^x)*2^(x-2)*lg3

We'll divide by 21 both
sides:

2^lg(10-3^x)*2^(x-2)*lg3 =
1

Since the bases are matching, we'll add the
superscripts:

2^[lg(10-3^x)+(x-2)*lg3] =
2^0

[lg(10-3^x)+(x-2)*lg3] =
0

We'll use the power
property:

[lg(10-3^x)+lg3^(x-2)] =
0

lg(10-3^x) =
-lg3^(x-2)

lg(10-3^x) =
lg3^-(x-2)

Since the bases are matching, we'll use one to
one property:

10-3^x =
3^-(x-2)

10-3^x = 9/3^x

We'll
substitute 3^x by t:

10 - t =
9/t

10t - t^2 - 9 = 0

t^2 -
10t + 9 = 0

t1 = 9 and t2 =
1

3^x = 9

x =
2

t2 = 1

x =
0

The requested values of x are {0 ;
2}.