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$2\log(a-2b) = \log((a-2b)^2)=\log(a)+\log(b) = \log(ab)\\ (a-2b)^2 = ab\\ a^2-4ab+4b^2 = ab\\ a^2 - 5ab + 4b^2 = 0\\ a = \dfrac{5b\pm\sqrt{25b^2-16b^2}}{2}\\ a = \dfrac{5b\pm 3b}{2} = b,4b\\ \text{however, }a\neq b \text{ as that would result in a negative number}\\ \text{as the argument to the log function in the original equation. So }\\ a=4b\\ \dfrac a b = \dfrac 1 4$

 5 

128    I used math papa

p.s. HAHAHA!!

I just found out the post I answered was posted 5 years ago its now 2019 XD

x ^(a^2-b^2) = x^(16)

So   a^2 - b^2 = 16      a^2 - b^2   is a difference of squares

         (a-b)(a+b) = 16        a+b is given as 2   sub this in

         (a-b) (2) = 16

         (a-b) = 8                      cool?  

No graphing needed.....    the x - intercept occurs when y = 0       sub that in to find the x value  (the y value is zero.....)

                                                y intercept occurs when x = 0         sub that in to find the corresponding y value (the y-intercept)....

  Go ahead...you can do it...  

Revised Question: How many different numbers between 1/1000 and 1000 can be written either as a power of 2 or as a power of 3, where the exponent is an integer?

It is good that you finally settled down and allowed yourself to learn something from EP

BUT

Well before you accepted that EP was correct you were certainly not polite to him.

At that point in time there was no thank you from you for the time EP put in, even if an answer is wrong you should still say 'thanks for your time.'

AND

At that point in time you also took points off EP.

I believe you would treat your classroom teacher with considerably more disrespect than you treated EP. 

So I do not think it would be inappropriate for your teacher to have punished you.

1 000 000 000

                    Here where the letter h starts

It's a good thing I'm learning from the best. If I was learning from my old math teacher I would have been kicked out and called a ##### ##### ##### ###### #### #### ####. A bunch of names.

Live and Learn! 

Thank you for sticking with me and my idiotness lol. IN THE END, WE TRIUMPHED!

for 1/2 I did .055 and .05 but I didn't do .5 which would be the right answer.

I TELL YOU IM AN IDIOT!!!!!!!!!!!!!!!!!!!!! 

Welllllll.....you need to check your calculator...or your calculator skills ....haha   

That isn't what i got.

The surface area of a rectangular prism is 2(l)(w)+2(l)(h)+2(w)(h). 

The surface area of a triangular prism is the area of the bases plus the lateral surface area !

Look at the possibilities I posted to your first question    here:

https://web2.0calc.com/questions/pre-calculus-trig_6

What do you think the answer is cupcake?

I mean which ones do you know are right and

which ones do you know are wrong

and which ones need investigation.

Yes, you are correct......2 is a prime number....I missed that one ! 

Thank you!! 

So, I'm right?

I'll use the first evalution to see if I get the same answer as you did:

Tan x/2 = +-  sqrt (   (1-cosx) /(1+cos x) )

              =+-sqrt (  (1- -3/5)/(1+ -3/5)  )   = + -  sqrt( 8/8  /  2/5)  = +-sqrt  (4)  =+2 or  -2       We know it is the negative one because it is in Q II

So YES!   The answer is  -2  !!!

Yes, two is prime, so the answer is 2*3*5*7=210.