What is the answer to ln(1/(x^2))+ln(x)=4 ?

The answer to ln(1/(x^2))+ln(x)=4 is x= 1/(e^4)

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ln(1/(x^2))+ln(x)=4

ln (\frac{1}{x ^{2}}) + ln (x) = 4

x = \frac{1}{e ^{4}}

Decimal Notation

x = 0.01831 \dots

Solution steps

ln (\frac{1}{x ^{2}}) + ln (x) = 4

Apply log rules

x = \frac{1}{e ^{4}}

Verify Solutions:

x = \frac{1}{e ^{4}}

True

The solution is

x = \frac{1}{e ^{4}}

ln (4 x) - 3 ln (x^{2}) = ln^{2} (x)

(ln (x))^{3} = 7 ln (x)

lo g_{4} (100 - 3 x) = 23219

6 = lo g_{10} (x)

x, - 5 lo g_{3} (2 x) = - 5

What is the answer to ln(1/(x^2))+ln(x)=4 ?
The answer to ln(1/(x^2))+ln(x)=4 is x= 1/(e^4)
Solve for x: ln(1/(x^2))+ln(x)=4
The solution is x= 1/(e^4)