logarithm function

[marinaa11]

Google says logarithm is a branch of Algebra, so I am seeking for help here
There is a task: a function is given, I have to find with what meanings of x f(x)≤0. Actually, there is a graph given, but the task says find meanings of x not using this graph.
My solution: I though that if I will write I will get x meanings, when the initial function is ≤0, but for ex.: if x=5, the meaning isn't ≤.
Could you, please, explain steps of solution?

 

[HallsofIvy]

Because of that "-" in front, f will be negative if and only if the logarithm is negative. Also, you should know that "logarithm" is the inverse of the "exponential". In particular, if \(\displaystyle y= log_{1/2}(x)\) then \(\displaystyle x= \left(\frac{1}{2}\right)^y\). So \(\displaystyle y= log_{1/2}(x)\) being positive is 1/2 to a positive power and that will less than 1. Here the "argument" of the logarithm is x+8 so you want 0<x+ 8< 1 or -8< x< -7.

 

[Dr.Peterson]

Google says logarithm is a branch of Algebra, so I am seeking for help here
There is a task: a function View attachment 22614 is given, I have to find with what meanings of x f(x)≤0. Actually, there is a graph View attachment 22615 given, but the task says find meanings of x not using this graph.
My solution: I though that if I will write View attachment 22616 I will get x meanings, when the initial function is ≤0, but for ex.: if x=5, the meaning isn't ≤.
Could you, please, explain steps of solution?

I think you are saying "meanings" where I would say "values". So you want to find the set of values of x for which [MATH]f(x)\le 0[/MATH].

The tricky part here is that when the base of a logarithm is less than 1, the log is a decreasing function; since it is here being multiplied by a negative number, your function f is an increasing function, as shown on the graph.

In your work, you first divided by -3 to obtain the correct inequality [MATH]\log_{\frac{1}{2}}(x+8)\ge 0[/MATH]. But when you then eliminate the log, in effect by raising the base, [MATH]\frac{1}{2}[/MATH], to the power on each side, this reverses the direction of the inequality, because [MATH]\left(\frac{1}{2}\right)^x[/MATH] is a decreasing function -- that is, as x increases, this decreases toward zero. So now you have [MATH]x+8\le 1[/MATH], and the solution is [MATH]x\le -7[/MATH]. However, x must also be in the domain of the function f, which is [MATH]x>-8[/MATH]. Combining these, the actual solution is [MATH]8<x\le -7[/MATH].

I'm not sure what you are saying at the end about x=5. I'm wondering if the work you show is not really your work, but perhaps corrected according to the provided answer, and you don't understand it. For x=5, we find that [MATH]f(5) = -3\log_{\frac{1}{2}}(x+8) = -3\log_{\frac{1}{2}}(5+8) \approx 11.1[/MATH], which shows it is not in the solution set. Did you think it should be?

 

[marinaa11]

I think you are saying "meanings" where I would say "values". So you want to find the set of values of x for which [MATH]f(x)\le 0[/MATH].

The tricky part here is that when the base of a logarithm is less than 1, the log is a decreasing function; since it is here being multiplied by a negative number, your function f is an increasing function, as shown on the graph.

In your work, you first divided by -3 to obtain the correct inequality [MATH]\log_{\frac{1}{2}}(x+8)\ge 0[/MATH]. But when you then eliminate the log, in effect by raising the base, [MATH]\frac{1}{2}[/MATH], to the power on each side, this reverses the direction of the inequality, because [MATH]\left(\frac{1}{2}\right)^x[/MATH] is a decreasing function -- that is, as x increases, this decreases toward zero. So now you have [MATH]x+8\le 1[/MATH], and the solution is [MATH]x\le -7[/MATH]. However, x must also be in the domain of the function f, which is [MATH]x>-8[/MATH]. Combining these, the actual solution is [MATH]8<x\le -7[/MATH].

I'm not sure what you are saying at the end about x=5. I'm wondering if the work you show is not really your work, but perhaps corrected according to the provided answer, and you don't understand it. For x=5, we find that [MATH]f(5) = -3\log_{\frac{1}{2}}(x+8) = -3\log_{\frac{1}{2}}(5+8) \approx 11.1[/MATH], which shows it is not in the solution set. Did you think it should be?

That was my work in the photo, but I made a mistake: as I got an answer x belongs from (-8;-7], but as a final answer, I wrote (8;-7] and when I tried to check are values (sorry for the incorrect initial word "meanings") of the function ≤0 , so I took 5 as it is between 8 and -7, but it had to be a number between -8 and -7, so I got a wrong value.
Thank a lot for answers and explanations.

 

[Dr.Peterson]

That was my work in the photo, but I made a mistake: as I got an answer x belongs from (-8;-7], but as a final answer, I wrote (8;-7] and when I tried to check are values (sorry for the incorrect initial word "meanings") of the function ≤0 , so I took 5 as it is between 8 and -7, but it had to be a number between -8 and -7, so I got a wrong value.
Thank a lot for answers and explanations.

Ah! I missed the positive 8. Of course, [MATH](8, -7][/MATH] is empty, as it means [MATH]8 < x \le -7[/MATH], not [MATH]-7\le x < 8[/MATH]. The number 5 is not in that interval! That's why I didn't understand your trying 5.