Can you help me solve this question?

[SucAtMath]

 

[Otis]

Hi S.A.M. Part of your post is not English. Can you translate? Otherwise, what have you tried, so far?

?

 

[canvas]

[math]\text{Just use basic exponent propertiess:}\,\,\,\,a^{bx}=(a^x)^b,\,\,\,a^x\cdot a^y=a^{x+y}\,\,\, \text{then put}\,\,\\3^x=t,\,\,t>0\,\,\,\text{and solve polynomial inequality, and show us your work.}[/math]

 

[canvas]

Hi S.A.M. Part of your post is not English. Can you translate? Otherwise, what have you tried, so far?

?

You don't have to translate it.. just looking on it: set A is given by the first inqeuality and set B is given by the second inequality, we need to find [math]A\cap B[/math]

 

[Otis]

You don't have to translate it …

Hi canvas. I agree with that statement!

 

[SucAtMath]

You don't have to translate it.. just looking on it: set A is given by the first inqeuality and set B is given by the second inequality, we need to find [math]A\cap B[/math]

The translation would be just like that, thanks!

 

[SucAtMath]

Hi S.A.M. Part of your post is not English. Can you translate? Otherwise, what have you tried, so far?

?

I‘ve tried and got results for both of them, but I‘m not sure if the answers are correct or not.

 

[topsquark]

I‘ve tried and got results for both of them, but I‘m not sure if the answers are correct or not.

So what are the results you got?

-Dan

 

[SucAtMath]

I got [1/2,2] for set A and (1/2,inf) for set B

 

[Eugenio]

For A you can consider [imath]t=3^x[/imath]. Then, the inequality is transformed into:

[math]t^3+27\leqslant 3t+9t^2[/math][math]t^3 -9t^2-3t+27\leqslant0[/math][math](t-9)(t^2-3)\leqslant0[/math]
That means [imath]t\epsilon [\sqrt{3};9)[/imath]

Finaly

[math]A=[1/2;2][/math]

 

[Otis]

I got [1/2,2] for set A and (1/2,inf) for set B

For A, I agree!

For B, please check your work. Using Change-of-Base formula on term log_x(8) would yield ln(8)/ln(x). That tells us there is at least one number in your solution set that you need to remove from the domain. Can you see what it is? Think about the graph of ln(x). Also, there is a short interval missing from your solution and one interval in there that produces values less than zero. We could confirm that by using a calculator to check some x-values on either side of that discontinuity in the domain I'd mentioned above.

Regarding your posted solution's parentheses, recall that the inequality is 'greater than or equal to zero'.

If you get stuck, consider showing some of your work. Maybe the mistake is something easy to fix.

?

 

[SucAtMath]

For A, I agree!

For B, please check your work. Using Change-of-Base formula on term log_x(8) would yield ln(8)/ln(x). That tells us there is at least one number in your solution set that you need to remove from the domain. Can you see what it is? Think about the graph of ln(x). Also, there is a short interval missing from your solution and one interval in there that produces values less than zero. We could confirm that by using a calculator to check some x-values on either side of that discontinuity in the domain I'd mentioned above.

Regarding your posted solution's parentheses, recall that the inequality is 'greater than or equal to zero'.

If you get stuck, consider showing some of your work. Maybe the mistake is something easy to fix.

?

Is the number 1?

 

[Infinity_lover]

Set A:[1/2 , 2]
Set B: ( - inf , 1/2] ∪ (1 , +inf)
Thus : A ∩ B = {1/2} ∪ (1,2]

 

[canvas]

[math]\log_ab=c \Leftrightarrow a^c=b\wedge a\neq1\wedge a>0\wedge b>0[/math]

 

[Otis]

Is the number 1?

Yes. The left-hand side of that inequality is not defined when x=1.

Another way to realize it is to consider the meaning of log_x(8). Logarithms are a way of expressing exponents.

x^? = 8

That unknown exponent may be expressed as a logarithm: log_x(8)

A concrete example could be letting the base equal 2. Then we'd have log₂(8)=3 because three copies of 2 multiplied together is 8.

2³ = 2*2*2 = 8

It does not make sense to let the base equal 1 because all powers of 1 yield 1. That is, there is no way to obtain the product 8 by multiplying copies of 1 together.

1¹ = 1
1² = 1*1 = 1
1³ = 1*1*1 = 1
1⁴ = 1*1*1*1 = 1

With logarithms using the Real number system, both the base and the argument must be positive numbers, and the base cannot be 1.

?

 

[Otis]

Set A:[1/2 , 2]
Set B: ( - inf , 1/2] ∪ (1 , +inf)
Thus : A ∩ B = {1/2} ∪ (1,2]

Hi I_L. Your set B may not contain negative numbers or zero.

[imath]\;[/imath]

 

[Infinity_lover]

Hi Otis. Sory,I can't understand what you mean.

 

[Otis]

… I can't understand what you mean.

That's okay; I'll say it another way.

log_x(a)

Symbol x is the base of the logarithm. It cannot be negative, zero or one. (Any positive Real number other than 1 is okay.)

Here's another way to see that x cannot be negative, in the given exercise. We have the logarithmic term:

log₂(√[2x])

√[2x] is not defined for negative x, in the Real number system.

?

 

[Infinity_lover]

That's okay; I'll say it another way.

log_x(a)

Symbol x is the base of the logarithm. It cannot be negative, zero or one. (Any positive Real number other than 1 is okay.)

Here's another way to see that x cannot be negative, in the given exercise. We have the logarithmic term:

log₂(√[2x])

√[2x] is not defined for negative x, in the Real number system.

?

Thank you so much for pointing out this important point that I forgot.
That's right.
Set B: (0 , 1/2] ∪ (1 , +inf)

 

[Infinity_lover]

Set B=(-inf , 1/2] ∪ (1 , +inf)