Question

In the given set $\left\{xϵR|\log \left[\right(1.6{)}^{1-x^2}-\left(0.625{)}^{6(1+x)}\right]ϵR\right\}$, $x$ belongs to

• A. $(-\infty ,-1)\cup (7,\infty )$
• B. $(-1,5)$
• C. $(1,7)$
• D. $(-1,7)$

Answer 1

The correct option is D $(-1,7)$

$log\left[\right(1.6{)}^{1-x^2}-\left(0.625{)}^{6(1+x)}\right]$

As we know that $lo{g}_{b}a$ is defined when $b>0,b\ne 1$ and $x>0$

Hence we have :
$(1.6{)}^{1-x^2}$ $>$ $(0.625{)}^{6(1+x)}$
${\frac{8}{5}}^{1-x^2}>$ ${\frac{8}{5}}^{-6-6x}$
$1-x^2>-6-6x$
$-x^2+7+6x>0$
$x^2-6x-7<0$
$(x+1)(x-7)<0$
$x$ belongs to $(-1,7)$