If x,y,andzare three consecutive positive integers, then logex+logez+12xz+1+1312xz+13+1512xz+15+..... is
logey
logey2
None of these
Explanation for the correct option.
Step 1: Solve 12xz+1+1312xz+13+1512xz+15+.....
We know that 12log1+a1-a=a+a33+a55+....
Here, let a=12xz+1
So,
12xz+1+1312xz+13+1512xz+15+.....=12log1+12xz+11-12xz+1=12log2xz+22xz=12logxz+1xz
Step 2: Solve logex+logez
logex+logez=logex1/2+logez1/2=12logex+12logez=12logxz
Step 3: Solve logex+logez+12xz+1+1312xz+13+1512xz+15+.....
logex+logez+12xz+1+1312xz+13+1512xz+15+.....=12logxz+12xz+1xz=12logxz×xz+1xz=12logxz+1
It is given that x,y,andzare three consecutive positive integers, x=y-1,andz=y+1.
12logxz+1=12logy-1y+1+1=12logy2-1+1=logy2×1/2=logey
Hence, option B is correct.
Find the value of x so that; (i) (34)2x+1=((34)3)3(ii) (25)3×(25)6=(25)3x(iii) (−15)20÷(−15)15=(−15)5x(iv) 116×(12)2=(12)3(x−2)