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Suppose that 4^a = 5, 5^b = 6, 6^c = 7, and 7^d = 8. What is a*b*c*d ?

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Suppose that 4^a = 5, 5^b = 6, 6^c = 7, and 7^d = 8. What is a*b*c*d ?  [#permalink]

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New post 24 Mar 2019, 23:27
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Question Stats:

45% (01:33) correct 55% (01:59) wrong based on 20 sessions

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Re: Suppose that 4^a = 5, 5^b = 6, 6^c = 7, and 7^d = 8. What is a*b*c*d ?  [#permalink]

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New post 25 Mar 2019, 00:09
It is given that 4^a=5, 5^b=6, 6^c=7 , 7^d=8

taking log on both sides , we get
alog4=log5 ==> a=log5/log4
blog5=log6 ==> b=log6/log5
clog6=log7 ==> c=log7/log6
dlog7=log8 ==> d=log8/log7


Thus,a*b*c*d = log8 /log4
= log(2)^3/log(2)^2
= 3/2

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Re: Suppose that 4^a = 5, 5^b = 6, 6^c = 7, and 7^d = 8. What is a*b*c*d ?  [#permalink]

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New post 25 Mar 2019, 00:10
1
Bunuel wrote:
Suppose that 4^a = 5, 5^b = 6, 6^c = 7, and 7^d = 8. What is a*b*c*d ?

(A) 1
(B) 3/2
(C) 2
(D) 5/2
(E) 3


7^d=8
so
8= (6^c)^d
8=((5^b)^c)^d
8= (((4^a)^b)^c)^d
so
2^3 = 2^2 a.b.c.d
so
3=2a.b.c.d
3/2 = a.b.c.d
IMO B
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Re: Suppose that 4^a = 5, 5^b = 6, 6^c = 7, and 7^d = 8. What is a*b*c*d ?  [#permalink]

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New post 26 Mar 2019, 00:16
2
given \(4^a\) = 5, \(5^b\) = 6,\(6^c\) = 7 , \(7^d\) = 8


Substitute 5 as \(4^a\) in \(5^b\) = 6 ==> \(4^{ab}\) =6
Similarly substitute for 6 and 7 we will get \(4^{abcd}\)= 8
===> \(2^{2abcd}\) = \(2^3\)
===> 2abcd = 3
abcd = \(\frac{3}{2}\)
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Re: Suppose that 4^a = 5, 5^b = 6, 6^c = 7, and 7^d = 8. What is a*b*c*d ?   [#permalink] 26 Mar 2019, 00:16
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Suppose that 4^a = 5, 5^b = 6, 6^c = 7, and 7^d = 8. What is a*b*c*d ?

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