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}$ ?


$ 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.

  • $\begingroup$ Applying $b>a^{k-1}$, we need to prove that $(k+4.2)(k-2)<4.2(k-1)(k-2)$.... how? $\endgroup$ – Mike SQ Sep 10 '18 at 9:48
  • $\begingroup$ It is just first-order inequality after cancelling out $k-2$. $\endgroup$ – didgogns Sep 10 '18 at 11:54
  • $\begingroup$ We need to prove, $ 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}$Note, $ a^{(k-2)(k+ 4.2)} \ll_k b^{(k-2)(k+ 4.2)}$ [ $\because b>a^{k-1}$] So,it is sufficient to prove -$b^{(k-2)(k+ 4.2)} \ll_k b^{8.4+2k} \implies (k-2)(k+ 4.2) \ll_k 8.4+2k$ how did u get $(k+4.2)(k-2)<4.2(k-1)(k-2)$? isn't it wrong? $\endgroup$ – Mike SQ Sep 12 '18 at 6:44

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