Plz help me asap plzzzzz..

Mathematics

1 answer:

Gre4nikov [31]3 years ago

6 0

28:

1*28 = 28

2*14 = 28

4*7 = 28

Factors of 28 are 1, 2, 4, 7, 14, 28.

15:

1*15 = 15

3*5 = 15

Factors of 15 are 1, 3, 5, 15.

36:

1*36 = 36

2*18 = 36

3*12 = 36

4*9 = 36

6*6 = 36

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36

29:

1*29 = 29

Factors of 29 are 1, 29.

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For what values of x is log base 0.8 (x+4)>log base 0.4 (x+4)<br>Thank you!

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We are seeking the solution of the inequality:

$\displaystyle{ \log_{0.8}(x+4)\ \textgreater \ \log_{0.4}(x+4)$ .

We recall that a log function $f(x)=\log_b(x)$ is either increasing or decreasing:

i) it is increasing if b>1,

ii) it is decreasing if 0<b<1.

Consider the functions $\displaystyle{ \log_{0.8}(x)$ and $\displaystyle{ \log_{0.4}(x)$ .

The graphs of these functions both meet at x=1 (clearly), and after 1 they are both negative. So from 0 to 1 one of them is larger for all x, and from 1 to infinity the other is larger. (Being strictly decreasing, their graphs can only intersect once.)

We can check for a certain convenient point, for example x=0.8:

$\displaystyle{ \log_{0.8}(0.8)=1$ and

$\displaystyle{ \log_{0.4}(0.8)=\log_{0.4}(0.4\cdot 2)=\log_{0.4}(0.4)+\log_{0.4}(\cdot 2)=1+\log_{0.4}(2)$ .

Now, $\displaystyle{ \log_{0.4}(2)$ is negative since we already explained that for x>1 both functions were negative. This means that

$\displaystyle{ \log_{0.8}(0.8)\ \textgreater \ \log_{0.4}(0.8)$ , and since $0.8\in (0, 1)$ , then this is the interval where $\displaystyle{ \log_{0.8}(x)\ \textgreater \ \log_{0.4}(x)$ .

So, now considering the functions $\displaystyle{ \log_{0.8}(x+4)$ and $\displaystyle{ \log_{0.4}(x+4)$ , we see that

x+4 must be in the interval (0,1), so we solve:

0<x+4<1, which yields -4<x<-3 after we subtract by 4.

Answer: (-4, -3). Attached is the graph generated using Desmos.