Logarithmic values and domains A real logarithm requires a positive argument. Its value may be positive, zero or negative. With u=ln A(x), the domain is A(x)>0, while u can be any real number. The logarithm is the exponent: ln A=u is equivalent to A= eu>0. A negative logarithm therefore represents a positive argument between zero and one. Find every argument restriction first. Combine logarithms only on that domain. Substitute the repeated logarithm; solve the quadratic; exponentiate each root; solve for x and verify the domain. Example For (ln x)2+ln x-2=0, require x>0. With u=ln x, (u-1)(u+2)=0; hence x= e or x= e-2. Both are valid. Check: do not reject u=-2. Reject a nonpositive logarithm argument instead. The identity ln(x2)=2ln x needs x>0 when the right side is used.

A real logarithm requires a positive argument. Its value may be positive, zero or negative. With \(u=\ln A(x)\), the domain is \(A(x)>0\), while \(u\) can be any real number.

The logarithm is the exponent: \(\ln A=u\) is equivalent to \(A=\mathrm e^u>0\). A negative logarithm therefore represents a positive argument between zero and one.

Find every argument restriction first. Combine logarithms only on that domain. Substitute the repeated logarithm; solve the quadratic; exponentiate each root; solve for \(x\) and verify the domain.

For \((\ln x)^2+\ln x-2=0\), require \(x>0\). With \(u=\ln x\), \((u-1)(u+2)=0\); hence \(x=\mathrm e\) or \(x=\mathrm e^{-2}\). Both are valid.

Check: do not reject \(u=-2\). Reject a nonpositive logarithm argument instead. The identity \(\ln(x^2)=2\ln x\) needs \(x>0\) when the right side is used.