In benzene molecule, carbon and carbon bond can't have polarity because there is not any electronegativity difference and the carbon hydrogen is also not much polar because the electronegativity difference is almost zero and also the small pull due to small difference in Electronegativity is balanced due to even pull from all directions.
In chlorobenzene the electronegativity difference between carbon and chlorine is very high so, it is polar.
(to compare polarity first check electronegativity...
Here, f(x)=x2 -6 logx-3=0
f(2)=4-6 log2-3=-0.806
f(3)=9-6 log3-3=3.1373
f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.
Hence, the root lies between 2 and 3
c0 =(2+3)/2=2.5
f(2.5)=6.25-6 log 2.5-3=0.8623
Now
n | a(-ve) | b(+ve) | cn | f(cn) |
0 | 2 | 3 | 2.5 | 0.8623 |
1 | 2 | 2.5 | 2.25 | -0.050595 |
2 | 2.25 | 2.5 | 2.375 | 0.38664 |
3 | 2.25 | 2.375 | 2.3125 | 0.1631658 |
4 | 2.25 | 2.3125 | 2.28125 | 0.05506 |
5 | 2.25 | 2.28125 | 2.265625 | 0.001925 |
From the table,
f(2.265625)=0.001928<10-2
Therefore, the...
Log2aa=x then, a=(2a)x ......(1)
Log3a2a=y then,2a=(3a)y ......(2)
Log4a 3a=z then, 3a=(4a)z ......(3)
So,
a=(2a)x [from (1)]
Or, a=(3a)xy [from(2)]
Or, a=(4a)xyz [from(3)]
Multiplying both sides by 4a,
4a.a=4a.(4a)xyz
Or,(2a)² =(4a)xyz + 1
Or,(3a)2y =(4a)xyz+1
Or,(4a)2yz =(4a)xyz+1
Or, 2yz = xyz+1 .proved.