Question

let3^a=4,4^b=5,5^c=6,6^d=7,7^e=8,8^f=9.The value of the product (abcdef),is

Answer 1

hi friend,

<>3^a=4

converting it into logarithm form,we get

log 4 base 3=a

[log a base b =loga/logb]

log4/log3=a------(1)

similarly, we get

log 5 /log 4=b----(2)

log6/log5=c-----(3)

log7/log6=d-----(4)

log8/log7=e----(5)

log9/log8=f----(6)

now abcdef = log 9/log3

[log a/log b=loga base b]

=log 9 base 3

=log 3² base 3

by power law,

=2 log 3 base 3

=2

I hope this will help u ;)

Answer 2

Hi ,

i ) 3^a = 4

3 = 4^1/a -------( 1 )

similarly,

ii) 4 = 5^1/b -------( 2 )

iii) 5 = 6^1/c ------( 3 )

iv )6= 7^1/d -------( 4 )

v) 7 = 8^1/e---------( 5 )

vi ) 8 = 9^1/f---------( 6 )

Now take ( 1 )

4^1/a = 3

( 5^1/b)^1/a = 3 [ from ( 2 ) ]

5^1/ab = 3

( 6^1/c)^1/ab = 3 [ from ( 3 ) ]

6^1/abc = 3

( 7^1/d)^1/abc = 3 [ from ( 4 ) ]

7^1/abcd = 3

( 8^1/e )^1/abcd = 3 [ from ( 5 ) ]

8^1/abcde = 3

( 9^1/f )^1/abcde = 3 [ from ( 6 ) ]

9^1/abcdef = 3

( 3^2 )^1/abcdef = 3

( 3 )^2/abcdef = 3^1

Therefore,

2/abcdef = 1 [ since if a^m = a^n then m = n ]

2 = abcdef

abcdef = 2

i hope this will useful to you.

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