Question

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

Answer 1

4^a = 5
a * log 4 = log 5
a = log 5 / log 4
a = 0.6989700043 / 0.6020599913
a = 1.1609640474

Answer 2

5^b = 6
b * log (5) = log (6)
b = log(6) / log(5)
b = 1.1132827526

Answer 3

c = 1.0860331325

Answer 4

d = 1.0686215613

Answer 5

\[\log_4(5)\times \log_5(6)\times\log_6(7)\times \log_7(8)\]

Answer 6

if you need to find a numeric answer, you can calculate each of these using the change of base formula

Answer 7

No need for numeric values if you observe the pattern

Answer 8

a * b * c * d =
1.1609640474 * 1.1132827526 * 1.0860331325 * 1.0686215613 =

1.5
Wow I didn't expect that

Answer 9

oops i think there is more to this problem than that
since the actual answer is \(\frac{3}{2}\)

Answer 10

@whpalmer4 what is the gimmick here?

Answer 11

Surprising that all those calculations reduce to 1.5

Answer 12

well, you have a = log[5]/log[4] b = log[6]/log[5] c = log[7]/log[6] d = log[8]/log[7]
multiply them all together and all the intermediate ones cancel out leaving abcd = log[8]/log[4] = 1.5

Answer 13

!!!

Answer 14

yeah, whodathunkit!

Answer 15

yes I didn't bother to look for a short cut, but who got the answer ?
:-)

Answer 16

you, apparently

Answer 17

I guess so - and it would have been MUCH easier to reduce all those calculations

Answer 18

think first, compute later :-)

Answer 19

I saw the two log 5/log 4 and log 6/log 5 things you posted and the lightbulb went on...

Answer 20

Yes it's like asking what is the sum of 1 + 2 + 3 all the way to 1,000
Yes you could add 1 + 2 + 3 + 4 ...
but think about it and n*(n+1) / 2 is a lot easier

Answer 21

If there were a way to give two medals... :P

Answer 22

it's the kind of problem which is just hard enough so that the brute force approach appears tractable, but takes long enough that the lazy guy might spot the pattern and look smart :-)

Answer 23

or as you said, it is the kind of problem where it seems that taking the long way is the right way