Question

Let $x={\log }_{4}9+{\log }_{9}28$ show that $\left[x\right]=3$, where $\left[x\right]$ denotes the greatest integer less than or equal to $x$. Find $\left[x\right]$.

Answer 1

${\log }_x{x}^n=n$

$x={\log }_{4}9+{\log }_{9}28$

So, ${\log }_{4}8+{\log }_{9}27<x<{\log }_{1}6+{\log }_{9}81$

$\Rightarrow \frac{3}{2}+\frac{3}{2}<x<2+2$

$\Rightarrow 3<x<4$

ie, $\left[x\right]=3$