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Inverse sine is the same as arc sine

So on this calc use    [2nd] [asin]

.

$$\\3log_2(x)-[log_2(3)-log_2(x+4)]\\\\
=3log_2(x)-log_2(3)+log_2(x+4)\\\\
=log_2(x^3)-log_2(3)+log_2(x+4)\\\\
=log_2\;\frac{x^3(x+4)}{3}\\\\$$

I am not sure if that is what you wanted :/

Yes, you said it  !!!     

My royal ego will explode :D

It is answered on the original thread - that is where the question is!

Heureka and I both answered it :)

YES that is true :)

Interusting...

Kitty has done it via formula.  There is nothing wrong with that.  Thanks Kitty.

CPhill hse done it very logically, he just forgot to mention that 2 of his isoscles triangle 'leg' angles adds up to one internal ocatgon angels.  His answer is the same as Kitty's.

I really like how you did that Chris.     

just to post it on here

10x10=100 obviously

Oh. Why did you do that?

i know i was seeing how long it would take someone to find the answer

First do 7x9 then 27/3 and then subtract! 

That IS in simplest form. You can't break it down any more.

Are three wise men enough to convince you Chris ?     :)

I have added this to the "Great Answers To Learn From" sticky thread    Thanks Guys 

No you can not :)

if TanA+1/TanA = 2 , find the value of tanA^2+1/tan^2A

$$\\tanA+\frac{1}{tanA} = 2\\\\
\left(tanA+\frac{1}{tanA}\right)^2 = 2^2\\\\
tan^2A+\frac{1}{tan^2A}+2*tanA*\frac{1}{tanA}= 4\\\\
tan^2A+\frac{1}{tan^2A}+2= 4\\\\
tan^2A+\frac{1}{tan^2A}= 2\\\\$$

Kitty has found the 2nd quadrant answer.

there is also a third quadrant answer and that would be

180+60 = 240 degrees 

or

4pi/3 radians

$${\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{10}} = {\mathtt{100}}$$

 $${{\mathtt{4}}}^{{\mathtt{12}}} = {\mathtt{16\,777\,216}}$$

$${\mathtt{16\,777\,216}}{\mathtt{\,\times\,}}{\mathtt{3.14}} = {\mathtt{52\,680\,458.24}}$$