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

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

Since log in an increasing function:

Then log 2 (x) > 2    iff   x > 4

Then the solution is :

x= ( 4, inf)

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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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

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sonu28 | Student, Grade 10 | eNotes Newbie

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log4x+log2x>3

1/2log2x+log2x>3

3/2(log2x)>3

log2x>2

4<x

x belongs to (4,+infinite)

sonu28's profile pic

sonu28 | Student, Grade 10 | eNotes Newbie

Posted on

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

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william1941 | College Teacher | (Level 3) Valedictorian

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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.

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neela | High School Teacher | (Level 3) Valedictorian

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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's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.).

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