You can put this solution on YOUR website! = 2 1/2
Note:In this problem the variable x can be assigned any positive value since
x is not going to feature in the final answer as you may see in the following steps
log[x](k).log[5](x) = 5/2 ----(1)
That is log[5](k) = 5/2 (using log[b](a)Xlog[c](b) = log[c](a) ---(*))
This implies k= (5)^(5/2)
(using definition log[b](N)=p implies and is implied by N = b^p
where N is a strictly positive number)
k= (5^2)X sqrt5
k = 25sqrt5
Answer: k= 25 times(root5)
Verification: k= (5)^(5/2) in
log[x](k).log[5](x) = 5/2 ----(1)
LHS = log[x](k).log[5](x)
={log[x][(5)^(5/2)]}.log[5](x)
={(5/2)log[x](5)} [log[5](x)]
=(5/2){log[x](5)X log[5](x)}
=(5/2)X1 {since log[b](a)=1/log[a](b) implying log[b](a)Xlog[a](b) = 1)
=5/2 = RHS
Detailed explanation:
log[x](k).{1/log[x](5)} = 5/2 (using log[b](a) = 1/log[a](b) )
log[x](k)/log[x](5) = 5/2
That is log[5](k) = 5/2
(using change of base rule: log[5](k) = log[x](k)/log[x](5)(the new base is x)
This implies k= (5)^(5/2)
(using definition log[b](N)=p implies and is implied by N = b^p
where N is a strictly positive number)
That is k= (5^5)^(1/2) (using (a)^(mn) = (a^m)^n )(here a = 5, m= 5,n= 1/2)
That is k= [5^(4+1)]^(1/2)
k = (5^4X5^1)^(1/2) (using a^(m+n) = a^mXa^n)
k = (5^4)^(1/2) X (5^1)^(1/2) (using (ab)^m= a^mXb^m )
k = 5^[4X(1/2)]X sqrt5
k= (5^2)sqrt5
k = (25)times sqrt5
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