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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.
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