_{2}x+log

_{4}x=2

Hi,

Not really sure how to solve this logarithm.

Log_{2}x+log_{4}x=2

Many thanks

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

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

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

Note that by the log conversion formula, Log_{4}x = Log_{2}x/Log_{2}4, so the problem becomes

log_{2}x(1 + 1/Log_{2}4) = 2, so (1.5)log_{2}x = 2 or log_{2}x = 2/(1.5) = 4/3,

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

x = cube root of 16.

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

Log_{2} X + log _{4 X = 2
}

log_{4} X= m X = 4^m = 2 ^(2m )

Log _{4 }X = 2 log_{2} X

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

Log (X .X^2)_{2 = }Log_{2} 4

X^3 = 4

X =^{ 3}√4

Parviz,

If you substitute X = 2^{(2/3)} in the original equation, you get 1 not 2. X is the cube root of 16, not the cube root of 4 as you stated.

Hi Aimee;

log_{2}x+log_{4}x=2

We need to work with the same base.

Let's convert one base.

I select the base of 4 because 2^{2}=4...

log_{4}x

(log_{2}x)/(log_{2}4)

(log_{2}x)/2

(1/2)log_{2}x

The equation is now...

log_{2}x+(1/2)log_{2}x=2

Let's combine like terms...

(3/2)log_{2}x=2

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

log_{2}x=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(2^{4})

x=cube root(16)

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