log2x+log4x=2

Question

Hi,

Not really sure how to solve this logarithm. 

 

Log₂x+log₄x=2

 

Many thanks

Answer 1

I would just add that x=2^(4/3)=2(2^(1/3))

so the simpler answer is x = 2 * cuberoot(2)

Answer 2

This problem becomes simple if you understand the meaning of logarithms, and recall some simple rules.

 

Note that by the log conversion formula, Log₄x = Log₂x/Log₂4, so the problem becomes

 

log₂x(1 + 1/Log₂4) = 2,  so (1.5)log₂x = 2 or  log₂x = 2/(1.5) = 4/3,  

 

Now remember, log₂x = 4/3 means "the power that you raise 2 to in order to get x is 4/3", so x = the cube root of 2⁴.  

 

x = cube root of 16.

 

 

 

Answer 3

 Log₂ X + log _{4 X = 2}

 

 

  log₄ X= m     X = 4^m = 2 ^(2m )

                      Log ₄ X = 2 log₂ X

 

 

    Log₂ X +Log _{4 X =} Log₂ X + 2 Log_{2 X =} Log ₂ X + Log ₂ X^2 = Log ₂ X^3 = Log ₂ 4

 

    Log (X .X^2)_{2 =} Log₂ 4

 

     X^3 = 4

 

    X = ³√4

Answer 4

Hi Aimee;

log₂x+log₄x=2

We need to work with the same base.

Let's convert one base.

I select the base of 4 because 2²=4...

log₄x

(log₂x)/(log₂4)

(log₂x)/2

(1/2)log₂x

The equation is now...

log₂x+(1/2)log₂x=2

Let's combine like terms...

(3/2)log₂x=2

Let's multiply both sides by 2/3...

log₂x=4/3

x=2^4/3

The position of 3 in the denominator of the exponential means this is a cube root...

x=cube root(2⁴)

x=cube root(16)