Hi Kumar,

We have,

$$16^{\frac{n}{2}} > 17^{19}$$, which can be rewritten as
$$\frac{16^{\frac{n}{2}}}{17^{19}} > 1$$
$$\frac{16^{\left(\frac{n}{2} - 19\right)} * 16^{19}}{17^{19}} > 1$$

Let $$\left(\frac{n}{2} - 19\right) = t$$, where $$2t$$ is an integer,
$$16^t * \left(\frac{16}{17}\right)^{19} > 1$$

Understand that, $$\left(\frac{16}{17}\right)^{19}$$ lies between $$0$$ and $$1$$, as long as this quantity is greater than $$\frac{1}{16^t}$$ or $$(0.0625)^t$$, the equality will always satisfy, we will have to check the same, in an exam setting, you can use calculator, here we'll use logarithm,

$$\log_{\frac{1}{16^t}}{\left(\frac{16}{17}\right)}> \frac{1}{19}$$
$$\frac{\log{\left(\frac{17}{16}\right)}}{t\log{16}} > \frac{1}{19}$$

We are looking for the smallest such value of $$t$$. We know that $$t$$ cannot be zero because $$\frac{n}{2}>19$$, so we'll try the next possible value for $$t$$ which is $$0.5$$, since this satisfies the inequality above, we'll get the value for $$n$$ as $$39$$.

Hope this helps.

Thanks🙏

You're welcome!