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
2. if a^x – b^y = c^z x, y, z are in GP. Prove that log base b of a = log base c of b
3. solve for x, 5^logx – 4^logx-1 = 4^logx – 5^logx-1
4.if 3 positive real numbers a, b and c, c+b is not=1, c-b is not=1 and log base c+b of a
plus log base c-b of a = 2 log base c+a of a plus log base c-a of a, then show that a and
b are two sides and c is the hypotenuse of a right angled triangle.
5. solve 2 log base 4 of x + 9log base x of 2 = 10
6. solve log 4 . log2 . log3 . (2^x + 17) = ½
7. find log base 6 of 16, if log base 12 of 27 = a
8. solve for x, 2log base x of a + log base ax of a + 3log base a^2x of a = 0
9. if log x/y-a = logy/z-x = logz/x-y then show that x^(y+z) . y^(z+x) . z^(x+y) = 1
10. solve 2 log base 9 of x + log base x of 27 + 4 = 0
thanks,
Tejas Koshy, Class XI
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! "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
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