Exponential inequality!

Algebra Level 3

4 ( 9 x ) 19 ( 3 x ) + 12 0 4(9^ x)-19(3^ x)+12\leq 0

For the above inequality to be true, x [ log b a , log b c ] x\in \left[ \log _{ b }{ a } , \log _{ b}{ c } \right] , where a , b a,b and c c are positive integers, find the minimum value of a a + b b + c c .


The answer is 7.75.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Samarpit Swain
Jan 27, 2015

Clearly, the left -hand side of the above inequality can be transformed into a quadratic with a simple substitution, y = 3 x y=3^x Therfore,

4 y 2 19 y + 12 0 4y^2 - 19y + 12\leq 0 => ( y 4 ) ( 4 y 3 ) 0 (y-4)(4y-3)\leq 0

Now, plotting the w a v y c u r v e wavy-curve for the above inequality we find that this inequality exists only when y y\in [ 3 / 4 3/4 , 4 4 ]

Replacing y = 3 x y=3^x again, we get:

3 x 4 3^x\leq4 a n d and 3 x 3 / 4 3^x\geq3/4

Therefore,

x l o g 3 4 x\leq log_{3}{4} a n d and x l o g 3 3 / 4 x\geq log_{3}{3/4}

HENCE,

x x\in [ l o g 3 3 / 4 log _{ 3 }{ 3/4 } , l o g 3 4 log _{ 3}{4} ] :)

That's exactly how I did it wow.

Kunal Verma - 6 years, 4 months ago

1 pending report

Vote up reports you agree with

×

Problem Loading...

Note Loading...

Set Loading...