Prime numbers are numbers that can only be divided by itself and 1. The largest possible prime number that fits the scenario is 113
Let the prime numbers be p1 and p2, where p1 > p2; and the odd numbers be x1 and x2
So, we have:
![p_1 + p_2 + x_1 + x_2 = 128](https://tex.z-dn.net/?f=p_1%20%2B%20p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128)
The largest prime number less than 128 is 127.
If
, then
becomes
![127 + p_2 + x_1 + x_2 = 128](https://tex.z-dn.net/?f=127%20%2B%20p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128)
![p_2 + x_1 + x_2 = 128-127](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128-127)
![p_2 + x_1 + x_2 = 1](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%201)
<em>This is not possible, because three positive integers cannot add up to 1</em>
The next largest prime number is 113
If
, then
becomes
![113 + p_2 + x_1 + x_2 = 128](https://tex.z-dn.net/?f=113%20%2B%20p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128)
Collect like terms
![p_2 + x_1 + x_2 = 128-113](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%20128-113)
![p_2 + x_1 + x_2 = 15](https://tex.z-dn.net/?f=p_2%20%2B%20x_1%20%2B%20x_2%20%3D%2015)
Let ![p_2 = 3](https://tex.z-dn.net/?f=p_2%20%3D%203)
becomes
![3 + x_1 + x_2 = 15](https://tex.z-dn.net/?f=3%20%2B%20x_1%20%2B%20x_2%20%3D%2015)
Collect like terms
![x_1 + x_2 = 15-3](https://tex.z-dn.net/?f=x_1%20%2B%20x_2%20%3D%2015-3)
![x_1 + x_2 = 12](https://tex.z-dn.net/?f=x_1%20%2B%20x_2%20%3D%2012)
and
are odd numbers.
So, we have:
![x_1 = 5\\x_2 =7](https://tex.z-dn.net/?f=x_1%20%3D%205%5C%5Cx_2%20%3D7)
This is true because
![x_1 + x_2 = 12](https://tex.z-dn.net/?f=x_1%20%2B%20x_2%20%3D%2012)
![5 + 7 = 12](https://tex.z-dn.net/?f=5%20%2B%207%20%3D%2012)
<em>Hence, the largest of the possible primes is 113</em>
Read more about prime numbers at:
brainly.com/question/4184435
Answer:
-9
Step-by-step explanation:
Answer:
-3
Step-by-step explanation:
Answer:
![x=25.24](https://tex.z-dn.net/?f=x%3D25.24)
Step-by-step explanation:
To find the value of
, we are solving the exponential equation
using logarithms.
Let's solve it step-by-step
Step 1. Divide both sides of the equation by -1 to get rid of the negative signs:
![-6.2^{0.1x} =-100](https://tex.z-dn.net/?f=-6.2%5E%7B0.1x%7D%20%3D-100)
![\frac{-6.2^{0.1x}}{-1} =\frac{-100}{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B-6.2%5E%7B0.1x%7D%7D%7B-1%7D%20%3D%5Cfrac%7B-100%7D%7B-1%7D)
![6.2^{0.1x} =100](https://tex.z-dn.net/?f=6.2%5E%7B0.1x%7D%20%3D100)
Step 2. Take natural logarithm to bot sides of the equation:
![ln(6.2^{0.1x)}=ln(100)](https://tex.z-dn.net/?f=ln%286.2%5E%7B0.1x%29%7D%3Dln%28100%29)
Step 3. Use the power rule for logarithms: ![ln(a^{x} )=xln(a)](https://tex.z-dn.net/?f=ln%28a%5E%7Bx%7D%20%29%3Dxln%28a%29)
For our equation:
and ![x=0.1x](https://tex.z-dn.net/?f=x%3D0.1x)
![ln(6.2^{0.1x)}=ln(100)](https://tex.z-dn.net/?f=ln%286.2%5E%7B0.1x%29%7D%3Dln%28100%29)
![0.1xln(6.2)=ln(100)](https://tex.z-dn.net/?f=0.1xln%286.2%29%3Dln%28100%29)
Step 4. Divide both sides of the equation by ![0.1ln(6.2)](https://tex.z-dn.net/?f=0.1ln%286.2%29)
![0.1xln(6.2)=ln(100)](https://tex.z-dn.net/?f=0.1xln%286.2%29%3Dln%28100%29)
![\frac{0.1xln(6.2)}{0.1ln(6.2)} =\frac{ln(100)}{0.1ln(6.2)}](https://tex.z-dn.net/?f=%5Cfrac%7B0.1xln%286.2%29%7D%7B0.1ln%286.2%29%7D%20%3D%5Cfrac%7Bln%28100%29%7D%7B0.1ln%286.2%29%7D)
![x=\frac{ln(100)}{0.1ln(6.2)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bln%28100%29%7D%7B0.1ln%286.2%29%7D)
![x=25.24](https://tex.z-dn.net/?f=x%3D25.24)
We can conclude that the value of x in our exponential equation is approximately 25.24.