# Solve the inequality: log 4 x + log 2 x > 3Solve the inequality: log 4 x + log 2 x > 3

hala718 | Certified Educator

log 4  (x) + log 2 (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/ 2 = (1/2)*log 2 x

==> log 4 (x) + log 2 (x)  > 3

==> (1/2) log 2 (x) + log 2 (x)  > 3

==> (3/2) log 2 (x)  > 3

Now multiply by 2/3

==> log 2 (x) > 2

We know that if log 2 (x) = 2  ==> x = 4

Then log 2 (x) > 2    iff   x > 4

Then the solution is :

x= ( 4, inf)

tonys538 | Student

The inequality `log_4 x + log_2 x > 3` has to be solved.

`log_4 x + log_2 x > 3`

= `log_(2^2) x + log_2 x > 3`

Use the property of logarithm `log_(b^c) a = (1/c)*log_b a`

= `(1/2)*log_2 x + log_2 x >3`

= `(log_2 x)*(3/2) >3`

= `(log_2 x) >2`

= `x > 2^2`

= x > 4

The inequality holds for all values of x greater than 4

sonu28 | Student

log4x+log2x>3

1/2log2x+log2x>3

3/2(log2x)>3

log2x>2

4<x

x belongs to (4,+infinite)

sonu28 | Student

We'll impose the constraint of existence of logarithms: x>0.

We'll create matching bases:

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

We'll re-write the inequality, all logarithms having the same base:

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 it is bigger than 1, so the function is increasing. If the function is an increasing function, then the direction of the inequality remains unchanged.

For log 2 (x) > log 2 (4)  =>  x>4

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

what is the need of this long method i can solve it only four line

william1941 | Student

Here we use the property of logarithms that log a + log b= log a*b. Also for log a, a has to be positive as the log of negative numbers is not defined.

log 4x + log 2x >3

=> log 4*2*x^2 >3

=> log 8 x^2 >3

3 is equal to log 10^3=log 1000.

Therefore 8 x^2>1000

=> x^2>1000/8

=> x^2 >125

Therefore x > sqrt(125) as x has to be positive.

neela | Student

log4x +log2 x > 3.

To solve for  x.

By property of logarithms, loga+logb = logab.

Therefore the given inequality becomex:

log (4x*2x) > 3 = log10^3.

log8x^2 > log1000. Taking anti log, we get:

8x^2 > 1000

x^2 > 1000/8 = 125.

x > sqrt(125) = 5sqrt5 .

Or

x < -sqrt125. Or x < - 5sqrt5 .

Check/Tally  for -sqrt125:

LHS = log4x+log2x = (log 4x*2x) = log8x^2 = log 8y^2 >3  where y <x < (-sqrt125). So y^2  >125. Or log 8*y^2 > log1000 = 3 = RHS.

log l

giorgiana1976 | Student

We'll impose the constraint of existence of logarithms: x>0.

We'll create matching bases:

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

We'll re-write the inequality, all logarithms having the same base:

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 it is bigger than 1, so the function is increasing. If the function is an increasing function, then the direction of the inequality remains unchanged.

For log 2 (x) > log 2 (4)  =>  x>4

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