In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that
that is, by (12), $$b^{k(k-2)} \ll a^{k(k-2)} (a^2b^2)^{2.1(k-2)} \ll_k b^{4.2(k-2)}b^{8.4+2k}$$
Equation (12) is- $$b \geq (k^ka^{a-2})^\frac{1}{2} $$
Here, $3 \leq k \leq 149, 2^{49} < a, 2< b .$
I couldn't figure out how $ a^{k(k-2)} (a^2b^2)^{2.1(k-2)} \ll_k b^{4.2(k-2)}b^{8.4+2k}$
So, how do we prove
$ a^{k(k-2)} (a^2b^2)^{2.1(k-2)} \ll_k b^{4.2(k-2)}b^{8.4+2k}$ ?
Edit:
$ a^{k(k-2)} (a^2b^2)^{2.1(k-2)} \ll_k b^{4.2(k-2)}b^{8.4+2k}$
$\implies a^{k(k-2)} (a^2)^{2.1(k-2)} (b^2)^{2.1(k-2)} \ll_k b^{4.2(k-2)}b^{8.4+2k}$
$ \implies a^{k(k-2)} a^{4.2(k-2)} b^{4.2(k-2)} \ll_k b^{4.2(k-2)}b^{8.4+2k}$
$ \implies a^{k(k-2)} a^{4.2(k-2)} \ll_k b^{8.4+2k}$
$ \implies a^{(k-2)(k+ 4.2)} \ll_k b^{8.4+2k}$
EDIT: Fixed a calculation mistake
From $a>k$, $k^k>1$ and $b \geq (k^ka^{a-2})^\frac{1}{2}$, $b\geq (a^{a-2})^\frac{1}{2}>(a^{k-2})^\frac{1}{2}=a^\frac{k-2}{2}$.
Since $b$ cancels out in your equation, it remains to prove that $a^{k(k-2)+4.2(k-2)}<b^{8.4+2k}$, or $a^{(k-2)(k+4.2)}<b^{2(k+4.2)}$. Applying $b>a^\frac{k-2}{2}$ immediately gives the desired result.