Question 1189461
<br>
{{{log(3,(1-x))>=log(3,(-x^2+x+16))}}}<br>
To start with, both the arguments must be positive:
{{{(1-x)>0}}} --> {{{x < 1}}}
{{{(-x^2+x+16)>0}}} --> {{{-3.53 < x < 4.53}}} (approximately)<br>
Putting those together, the possible approximate range of solutions is<br>
[1] {{{-3.53 < x < 5.11}}}.<br>
Now we need to find which values of x in that range satisfy the given inequality.<br>
{{{log(3,(1-x))>=log(3,(-x^2+x+16))}}}<br>
Since log(A) is monotonic...<br>
{{{1-x>=-x^2+x+16}}}<br>
{{{x^2-2x-15>=0}}}<br>
{{{(x-5)(x+3)>=0}}}<br>
[2] {{{x>5}}} or {{{x<-3}}}<br>
Combining [1] and [2], we see that the inequality is satisfied only on the interval from -3.53 (approximately) to -3.<br>
ANSWER: {{{-3.53<x<-3}}}<br>