log(x,2log(2x, 2log(2,4x))) >1 , find answers of x

Asked May 01 '16 at 15:39

Active May 01 '16 at 15:56

Viewed 161 times

how i slove it , please help me?

log(x,2log(2x, 2log(2,4x))) >1

my try:

if x>1

=>2log(2x, 2log(2,4x))>x

=> 2log(2,4x)>(2x)^(x/2)

=>4x>2^(((2x)^(x/2))/2)

another way

log(x,2log(2x, 2log(2,4x)))=log(x,log(2x, (log(2,(4x)^2))^2))

but all of them is incorrect

please help me

i need it

it's so hard i think

edited May 01 '16 at 15:56

asked May 01 '16 at 15:39

phpcode

What do you mean by $\log (2, 4x)$? Also check out the mathJax tutorial :) – Riccardo Orlando May 01 '16 at 17:43
log(2,4x) means (log4x)/(log2) – phpcode May 04 '16 at 11:31
So log(a, b) = $\log_a(b)$? – Riccardo Orlando May 04 '16 at 11:34
yes exactly
yes yes yes – phpcode May 04 '16 at 11:42