this is because the expression within the log sign must be positive.
you get:
x < 1 and x > -3.5311 and x < 4.5311.
since x < 1 is the controlling influence on the x < side of the inequality, you get x > -3.5311 and x < 1
the admissible range is therefore -3.5311 < x < 1
within that range, you need to find out when log3(1-x) >= log3(-x^2 + x + 16).
if you set them equal to each other, you get:
log3(1-x) = log3(-x^2 + x + 16)
this is true if and only if 1-x = -x^2 + x + 16
subtract (1-x) from both sides of this equation to get:
0 = -x^2 + x + 16 -(1-x)
simplify to get:
0 = -x^2 + 2x + 15
multiply both sides of this equation by -1 to get:
0 = x^2 - 2x - 15]
factor this equation to get:
0 = (x-5) * (x+3)
solve for x to get:
x = 5 or x = -3
x can't be 5 since the permissible for the log function is x < 1.
x can be -3 since that is within the permissible range of x.
you have 1-x = -x^2 + x + 16 when x = -3.
this means that log3(1-x) = log(-x^2 + x + 16 when x = -3.
this means that log3(1-x) - log(-x^2 + x + 16 = 0 when x = -3.
log3(1-x) will be greater than log3(-x^2 + x + 16) when x is < -3 or when x is > -3
at his point, you just need to evaluate when x is less than -3 and when x is greater than -3.
you can use the log base conversion formula to test this out using a scientific calculator.
the log conversion formula is log3(x) = log(x)/log(3)
i used x = -3.5 for x less than -3 and x = .5 for x greater than -3 because x can't be greater than or equal to 1 and x can't be less than or equal to -3.5311 as we found out earlier.
when x = -3.5:
log(1-x) = 1.369070246.
log(-x^2 + x + 16) = -1.261859507.
this means that log(1-x)/log(3) is greater than log(-x^2 + x + 16)/log(30 when x is less than -3 and greater than -3.5311
when x = .5:
log(1-x) = -.6309297536.
log(-x^2 + x + 16) = 2.537831533.
this means that log(1-x)/log(3) is less than log(-x^2 + x + 16)/log(3) when x is greater than -3 and less than x = 1.
your solution should be that log3(1-x) >= log3(x + 16 - x^2) when x is greater than -3.5311 and x is less than or equal to -3.
i confess that i didn't figure this out until i graphed the function.
i used the desmos.com calculator.
it can be found at https://www.desmos.com/calculator
the graph looks like this.
i used a quadratic equation solver to find the 0 points for x + 16 - x^2.
the 0 point of interest was at x = -3.5311288741493.
x had to be greater than that for log3(x + 16 - x^2) to be valid.
consequently, the permissible value of x became:
x > -3.5311288741493 and x <= -3.
when x = -3, log3(1-x) = log3(x + 16 - x^2)
when x < -3 and > -3.5311288741493, log3(1-x) > log3(x + 16 - x^2).
your solution is that log3(1-x) >= log(x + 16 - x^2) when x > -3.5311288741493 and when x <= -3.
if you use the graph solution, you can say that x > -3.531 x <= -3.
this translates to -3.531 < x <= -3.
tough problem.
i'm not sure how much rounding they allow, but you can adjust from the unrounded figures as necessary.
the first thing to do was find the permissible range of log3(1-x) and log3(x + 16 - x^2).
that told you the that the value of x had to be greater -3.5311.... and less than 1.
the next thing to do was find out when they were equal.
that led to x = -3.
the next thing to do was find out when log3(1-x) was greater than log3(x + 16 - x^2) within the permissible range of x.
that's when you evaluated in the range of x > -3.5311 and x < -3 and the range of x > -3 and x < 1.
being able to graph the equation of y = log3(1-x) - log3(x + 16 - x^2) was a big help to visualize where the solution would lie.
let me know if you have any questions.
theo