Question 310922
"my earlier submission looks to be jumbled. I have tried to put the same questions in a better way. I am a beginner in logarithms. Please help me to find the solution for the below problems."
1. P = log base 12 of 18 Q = log base 24 of 54 show that PQ + 5(P-Q)=1
log12(12)=log12(3)+log12(2^2)=log12(3)+2log12(2)=1
log24(24)=log24(3)+log24(2^3)=log24(3)+3log24(2)=1
logb(mn)=logb(m)+logb(n)
conversion from base b to base k --> logb(x)=logk(x)/logk(b)
P=log12(18)=log12(2*3^2)=log12(2)+2log12(3)
Q=log24(54)=log24(2*3^3)=log24(2)+3log24(3)
   log12(2)log24(2)
+ 3log12(2)log24(3)
+ 2log12(3)log24(2)
+ 6log12(3)log24(3)
+ 5log12(2)(log24(3)+3log24(2))
+10log12(3)(log24(3)+3log24(2))
- 5log24(2)(log12(3)+2log12(2))
-15log24(3)(log12(3)+2log12(2))
-->
   log12(2)log24(2)
+ 3log12(2)log24(3)
+ 2log12(3)log24(2)
+ 6log12(3)log24(3)
+ 5log12(2)log24(3)
+15log12(2)log24(2)
+10log12(3)log24(3)
+30log12(3)log24(2)
- 5log12(3)log24(2)
-10log12(2)log24(2)
-15log12(3)log24(3)
-30log12(2)log24(3)
-->
   log12(2)log24(2)
+15log12(2)log24(2)
-10log12(2)log24(2)
+ 3log12(2)log24(3)
+ 5log12(2)log24(3)
-30log12(2)log24(3)
+ 2log12(3)log24(2)
+30log12(3)log24(2)
- 5log12(3)log24(2)
+ 6log12(3)log24(3)
+10log12(3)log24(3)
-15log12(3)log24(3)
-->
  6log12(2)log24(2)
-22log12(2)log24(3)
+27log12(3)log24(2)
+  log12(3)log24(3)
-->
  6log2(2)*1/log2(12)*log2(2)*1/log2(24)
-22log2(2)*1/log2(12)*log2(3)*1/log2(24)
+27log2(3)*1/log2(12)*log2(2)*l/log2(24)
+  log2(3)*1/log2(12)*log2(3)*1/log2(24)
-->
                6/(log2(12)log2(24))
-       22log2(3)/(log2(12)log2(24))
+       27log2(3)/(log2(12)log2(24))
+  log2(3)log2(3)/(log2(12)log2(24))
-->
6 + 5log2(3) + log2(3)log2(3)
-->
log2(12)log2(24)
log2(3*4)log2(2*3*4)
(log2(3)+log2(4))(log2(2)+log2(3)+log2(4))
(2+log2(3))(3+log2(3))
6 + 2log2(3) + 3log2(3) + log2(3)log2(3)
6 + 5log2(3) + log2(3)log2(3)
-->
(6 + 5log2(3) + log2(3)log2(3))/(6 + 5log2(3) + log2(3)log2(3))
1/1
1
DONE
problems 2-10, I do not have time to do, took me even a long time to do problem 1, best of luck