IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

mihirathwa

Value of x when

5²5⁴5⁶.........5^2x= (0.04)^-28

johri_anshuman

5²5⁴5⁶.........5^2x
=5^{2(1+2+3......x)}
=5^x(x+1)Now,

5^x(x+1)=(0.04)^-28taking log on both sides
=>x(x+1)ln5=-28(ln(4*10^-2))

=>x(x+1)ln5=56ln4
=>x(x+1)=48.235
=>x²+x-48.235=0

x=(-1+-sqrt(48.235))/2

savvej

0.04=4/100=1/25=1/5²=5^-2
(0.04)^-28=(5^-2)^-28=5⁵⁶
5² 5⁴ 5⁶.........5^2x
=5²^{(1+2+3......x)}
=5^x(^x+1)
.: 56=x(x+1)
x2+x-56=0
.: x+8x-7x-56=0
.:(x+8)(x-7)=0
.:x=7 ,x=-8

krishna.gopal

Good savvej but you could have done it fast if you can see that 0.04 is 5^-2 So RHS becomes 5⁵⁶

Or x(x+1) = 56

Out of two answer we can only take x=7 because the way this expression is given we cannont take negative values

spideyunlimited

superb krishnagopal... it looks easy now :)

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