IIT JEE , AIEEE Forum - Mathematics , Algebra

dixit_se

solve
log₃(3^x-8)=2-x

and

18^8-4x=(54

2)^3x/2

nivedh_89

answer1)

therefore,
3^x - 8 = 3^(2-x)
= 3^2/3^x
= 9/3^x
therefore

3^2x-8*3^x-9 = 0
let,3^x = y

hence
y^2-8y-9 = 0
(y+1)(y-9) = 0
y= -1 or y=9
as 3^x cannot b negative
3^x = 9
x = 2...............!!!!!!!!!!!!!

spideyunlimited

ans 1) 3 ^ (2-x) = 3^x - 8
8 = 3 ^ x - 3 ^ ( 2-x)

as we can write 8 = 9 - 1
thus, we get x = 2.

nivedh_89

answer2)
as 54 = 18*3,

18^(8-4x) = (18^(3x/2))*(3root2)^(3x/2)
and 18 = (3root2)^2

so,

(3root2)^(16-8x-3x) = (3root2)^(3x/2)

therefore,
solving,
x = 32/25...............!!!!!!!!!!!!

spideyunlimited

yes nivedh is correct :))

spideyunlimited

in such questions try to find the relation between the given numbers. it helps a lot

metal

Excellent Nivedh

metal

Thanks for the really useful tip spidey

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