# How to solve 2ln3x = 2+ln 4? I don't understand how to that.

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I interpret this as **2*ln(3x) = 2+ln(4)**.

Note that x must be >0 for ln(3x) to make sense.

We want to move everything under the logarithm and then eliminate it.

To move a factor under the logarithm, we use ln(a^b) = b*ln(a):

2*ln(3x) = ln((3x)^2).

To move a summand under the logarithm, we use the previous and ln(a*b) = ln(a) + ln(b):

2 + ln(4) = 2*ln(e) + ln(4) = ln(e^2) + ln(4) = ln(4*e^2).

Now remember that 2*ln(3x) = 2+ln(4), or now

ln((3x)^2) = ln(4*e^2).

ln is one-to-one function, so ln(a)=ln(b) means that a=b:

(3x)^2 = 4*e^2.

Now we can extract positive square root because x>0:

3x = sqrt(4*e^2) = 2*e.

So the answer is **x = (2/3)e**.

`2ln3x=2+ln4`

`2ln3x-ln2^2=2`

`2ln3x-2ln2=2`

`ln3x-ln2=1`

`ln((3x)/2)=1`

`(3x)/2=e`

`:.x=(2e)/3`

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