log 2 (x) + log 4 (x) >3

We know that: log a b = log c b / log c a

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

==>log 2 (x) + log 4 (x) = log 2 (x) + log 2 (x)/ log 2 (4) > 3

But log 2 (4) = 2

==> log 2 (x) + log 2 (x) /2 > 3

==> (3/2) log 2 (x) > 3

Multiply bu 2/3

==> log 2 (x) > 2

But if log 2 (x) = 2 ==> x = 4

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

First, we'll express log 4 x = log 2 x / log 2 4

But log 2 4 = log 2 (2^2) = 2*log 2 2 = 2*1 = 2

log 4 (x) = log 2 (x) / 2

The inequality will become:

log 2 (x) + log 2 (x)/2 >3

We'll calculate the LCD and we'll add the terms:

3*log 2 (x) / 2>3

We'll divide by 3 both sides:

log 2 (x) / 2 > 1

We'll cross multiply:

log 2 (x) > 2

But 2 = log 2 (4)

log 2 (x) > log 2 (4)

The logarithms have the same base and is bigger than 1, so the function is increasing.

If log 2 (x) > log 2 (4), then x>4

**x belongs to the interval (4, +inf.).**

To solve the inequality log2 (x)+log4 (x) > 3

We know that log 4 (x) = 2 log2 (x) and 3 = log2 (2^3)

Therefore

LHS : log 2 (x) +2 log2 (x) = 3 log2 (x) Or log2 (x^3),

RHS: log2( 2^3.)

Therefore log2 (x^3 )> log2( 2^3). Taking antilogarithms,

x > 3.