Question

Solve the following inequality:
${\log }_{\pi }(x+27)-{\log }_{\pi }(16-2x)<{\log }_{\pi }x$

Answer 1

The following conditions must be checked to satisfy logarithm:

$(x+27)>0x\in (-27,\infty )$ $16-2x>0x\in (-\infty ,8)$ $x>0x\in (0,\infty )$

Taking intersection, we get $x\in (0,8)$

${\log }_{\pi }(x+27)-{\log }_{\pi }(16-2x)<{\log }_{\pi }x{\log }_{\pi }\frac{x+27}{16-2x}<{\log }_{\pi }x\frac{x+27}{16-2x}<x$

For $x\in (0,8)16-2x>0$

$x+27<x(16-2x)x+27<16x-2x^{2}2x^2-15x+27<0(2x-9)(x-3)<0x\in (3,4.5)$

Hence, $x\in (3,4.5)$