CPhill

So...we really have this.....√81 / √7 = 9 / √7

Multiplying the numerator and denominator by √7, we obtain

[9 √7] / [7]

And that's it.....

JF = KJ  = 10     Sides opposite equal angles are themselves equal 

Therefore, m<KJF = 120 

m<KJF = 120   Therefore m< KJA = 60   (The sum of any two angles comprising a strainght angle = 180)

Therefore, triangle KAJ  is 30 - 60 - 90

And AJ = 5     And KA = 5√3

Then, by the Pytagorean theorem,  KF = √[ (AJ + JF)^2 + (KA)^2] = √[ (5 + 10)^2 + (5√3)^2] =

 √[ (15)^2 + (5√3)^2] =  √[ 15^2 + 75] =  √[ 225 + 75]  = √[ 300] = 10√3

So....the perimeter =  AJ + JF + KF + KA = 5 + 10 + 10√3  + 5√3 =

15 + 15√3

The only thing we would need to check here is if anything made the denominator = 0.

However, a quick look tells us that this can't happen because 3x^2 is always positive (or, 0) and when I add 12 to a positive - or 0- I still have a positive number. Hence, no "x" valuies make the denominator = 0.

77,040

Here's an easy way to solve this type of problem.

Just add 1/16 and 1/12 = 28/192

Now...."flip" the fraction over

192/28 = 6.86 hrs.   !!!

And there you go.....

I may be incorrect about this, Melody, but I believe that x must be greater than 8. The answer has to make sense in the original problem, as well. If x is only greater than -8, the second expression in the original problem is undefined for (-8,8].

There is a way to get an exact answer, but it's not obvious, at first......

Note that we can write 81 as 3⁴ and we can write 256 as 4⁴

And using a log property, we can write

log₁₂ 81 + log₁₂ 256      .....as......

log₁₂ 3⁴ + log ₁₂ 4⁴ =  log₁₂(3⁴ * 4⁴) = log₁₂(3*4)⁴ = log₁₂ (12)⁴ = 4log₁₂(12) = 4*1 = 4

And that's it.......

I'm not sure how to solve this algebraically......so let's do the next best thing.....graph it!!

I'm subtracting 5^x from both sides, so we have

6^x - 5^x = 8

Here's the graph of the left side and the right side functions....the intersection point is about  x = 1.857

We have a rule here that helps on this one......

log a - log b  =  log(a/b)   Note that the bases have to be the same!!

So we have

log₃ 54 - log₃ 2   =  log₃ (54/2) = log₃ 27

And remember, we're just asking, "What power do we put on 3 to raise 3 to 27??"

The answer is.........3 !!!!

This one is similar to the one I just worked

Basically, 6-3x has to be greater than 0   ...   so we have

6-3x > 0     Add 3x to both sides

6 > 3x        Divide both sides by 3

2 > x      which means the same thing as x < 2

So our domain is

(-∞, 2)

And that's it......