Question

Find the least value of the expression \(2\log_{10} x - \log_x {0.01} \) for x >1

Answer 1

@hartnn @Miracrown

Answer 2

For this question,

\(2\log_{10} x + 2\log_x {10} \)

oh..

Answer 3

that log has base 10, right ?

Answer 4

\(2\cfrac{1}{\log_x {10} } + 2\log _x {10}\)

\(2 \left( \cfrac{1 + (\log_x {10})^2}{\log_x {10}}\right)\)

Answer 5

Yep.. its 10.

Answer 6

I don't : whether I am on right track here , if yes, then ho wto continue?

Answer 7

and you cannot use calculus ?

Answer 8

Oh... maxima minima?

Answer 9

least value means minima

Answer 10

Okay right.. so, if I differentiate it and equate to zero

Answer 11

if you can use calculus

Answer 12

u = log x

2(u+1/u)
diff. w.r.t u
equate to 0

Answer 13

Oh.. k i will try now.

Answer 14

\(\cfrac{d}{du} 2(u + \cfrac{1}{u}) \)

= \( 2( 1 + (-1) \cfrac{1}{u^2})\) = 0

1 = 1/u^2

u^2 = 1

Answer 15

2(u + 1/u ) = 2 (1+1/1) = 2(2) = 4

Answer 16

2(1-1/u^2) -> differentiating it again and putting u = 1

we get : 4 / u^3

put u = 1

so, 4/1 = positive

so, u = 1 is the minima , right @hartnn ?

Answer 17

O.O I just got to know another soln for this : (courtsey @vishweshshrimali5 )

\((u + \cfrac{1}{u} ) \ge 2 \)

So, 2(u+1/u) \(\ge\) 4

thus, 4 was the answer...

Answer 18

o.O it was much easier with the second method... but, yep, I always like calculus methods @hartnn ;)

Answer 19

nice!