SOLUTION: Compare 10^2 and 2^10. For any two numbers make a conjucture about which usually gives the greater number, as the base or as the exponent.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Compare 10^2 and 2^10. For any two numbers make a conjucture about which usually gives the greater number, as the base or as the exponent.      Log On


   

Question 1118244: Compare 10^2 and 2^10. For any two numbers make a conjucture about which usually gives the greater number, as the base or as the exponent.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
10%5E2 versus 2%5E10


What if both were to be in base 2?

2%5Ep=10%5E2
log%282%2C2%5Ep%29=log%282%2C10%5E2%29
p%2A1=2%2Alog%282%2C10%29

2%5E%282%2Alog%282%2C10%29%29 versus 2%5E10

Compare 2%2Alog%282%2C10%29 with 10.

Note that 2^3=8 and 2^4=16 so 3%3Clog%282%2C10%29%3C4.
Knowing that log(2,10)<=4, setup the inequality 2%2A4%3C10.


2%2Alog%282%2C10%29%3C10
--
10%5E2%3C2%5E10

Answer by ikleyn(52903) About Me  (Show Source):
You can put this solution on YOUR website!
.
Compare 10^2 and 2^10. For any two numbers make a conjucture about which usually gives the greater number, as the base or as the exponent.
~~~~~~~~~~~~~~~~~~


I'd like to answer the second question of this post. 

    For any two numbers make a conjecture about which usually gives the greater number, as the base or as the exponent.


So,  the question is:   if  x  and  y  are two positive numbers and   y > x,  what is greater,   x%5Ey  or  y%5Ex ?

The answer is:  if both  x  and  y  are greater than  "e"  and   x < y,  then   x%5Ey > y%5Ex.           (*) 

    In wording form: if in the domain  x, y > e  you raise the smaller number to the greater degree, you will get the greater number 
    than if you raise the greater number to the smaller degree.


Solution 

Let's start from this inequality

    x%5Ey > y%5Ex.               (1)


Take the logarithm from both sides. You will get


    y*ln(x) > x*ln(y).      (2)

Divide both sides by the product  ln(x)*ln(y).  You will get

    y%2Fln%28y%29 > x%2Fln%28x%29.            (3)           <<<---=== in the domain  x, y > e you are safe to do it . . . 


Now the KEY STATEMENT is that the inequality (3) is TRUE (!).    


Why it is true ?  - Because the function  F(x) = x%2Fln%28x%29  monotonically increases as  x  increases in the domain  x > e.


    I will not prove the last statement.


    Everybody who is familiar with Calculus, can easily do it on his or her own.


    Which is better, I will demonstrate this fact visually, by presenting the plot:



    graph%28+330%2C+330%2C+-1.5%2C+10.5%2C+-5.5%2C+10.5%2C%0D%0A++++++++++x%2Fln%28x%29%0D%0A%29


    Plot y = ln%28x%29%2Fx


Now, if you reverse the arguments in this sequence  (3)  ---->  (2)  ---->  (1), you will get the major statement  (*)  proved.

-----------------

So,  to the popular question - which number is greater,   e%5Epi  or   pi%5Ee,   the answer is:   e%5Epi   is greater than   pi%5Ee.