log 2 x - log 4 x = 2 find x

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log2x-log4x = 2.

Therefore

log(2) x - log(4) x = 2

We convert log(4)x to log2(x)/log(2)4 to ve the same base.

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

log(2)x -(1/2)log(2)x = 2

(1-1/2) logx = 2

(1/2) log(2)x = 2

log(2)x = 4

x = 2^4 = 16

If the base of logarithms is 10 ,so the bases are matching, we'll use the quotient property:

log 2 x - log 4 x = 2

log 2x/4x = 2

We'll simplify:

log 1/2 = 2

1/2 = 10^2 impossible for any value of x!

But if the bases are 2 and 4, we'll solve the equation:

log 2 x = log 4 x*log 4 2

log 2 x = log 4 x / log 2 4

log 2 x = log 4 x/2

log 4 x = 2 log 2 x

We'll use the power property of logarithms:

log 4 x = log 2 x^2

We'll substitute log 4 x by the log 2 x^2.

log 2 x - log 2 x^2 = 2

We'll use the quotient property:

log 2 x/x^2 = 2

log 2 1/x = 2

1/x = 2^2

1/x = 4

**x = 1/4**

**Since the solution is positive, it is accepted.**

The equation `log_2 x - log_ 4 x = 2` has to be solved for x.

First convert all the logarithm to a common base. As the numbers here are 2 and powers to two we use the base 2. Use the following property useful when changing the base of logarithms.

`(log_b x) = (log_n x)/(log_n b)`

`log_2 x - log_ 4 x = 2`

`log_2 x - (log_2 x)/(log_2 4) = 2`

`log_2 x - (log_2 x)/(log_2 2^2) = 2`

use the property `log a^b = b*log a`

`log_2 x - (log_2 x)/(2*log_2 2) = 2`

Now `log_b b = 1`

`log_2 x - (log_2 x)/(2*1) = 2`

`(1/2)*log_2 x = 2`

`log_2 x = 4`

If `log_b x = y` , `x = b^y`

`x = 4^2`

x = 16

The required solution is x = 16

log 2 x - log 4 x = 2

Let us rewirte:

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 2 x - (1/2) log 2 (x) = 2

==> (1/2) log 2 (x) = 2

No multiply by 2:

==> log 2 (x) = 4

==> x= 2^4

**==> x= 16 **

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