Mathematics, 28.07.2019 10:10 antonio9768

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

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Answer from: carog24

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.

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

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

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

Answer from: Quest

step-by-step explanation:

Answer from: Quest

starting with:

5 + x - 12 = 2x - 7

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)

5 - 12 = x - 7 (combine and cancel x's from both sides)

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)

0 = x (add numbers together on both sides)

so in order for the equation to be true, x must be 0.

trying x = -0.5

5 + x - 12 = 2x - 7 (original equation)

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)

5 + -0.5 - 12 = -1 - 7 (did multiplication)

-7.5 = -8 (did addition on both sides)

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.

trying x = 0

5 + x - 12 = 2x - 7 (original equation)

5 + 0 - 12 = 2(0) - 7 (plug in value of x)

5 + 0 - 12 = 0 - 7 (did multiplication)

-7 = -7 (did addition on both sides)

since -7 equals -7, that demonstrates that the value of 0 for x is correct.

trying x = 1

5 + x - 12 = 2x - 7 (original equation)

5 + 1 - 12 = 2(1) - 7 (plug in value of x)

5 + 1 - 12 = 2 - 7 (did multiplication)

-6 = -5 (did addition on both sides)

since -6 is obviously not equal to -5, the value of 1 for x is wrong.

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