From an Implicit Relation to an Explicit Formula
An implicit relation describes how \(x\) and \(y\) are connected without isolating \(y\). For example, \(\ln|y|=2x+C\) relates the variables, but it does not yet state \(y\) as a formula in \(x\). Solving explicitly means rearranging the relation into one or more formulas of the form \(y=\phi(x)\).
This step requires care. Logarithms of absolute values can lead to both positive and negative possibilities, while algebraic equations can have multiple roots. An explicit formula must represent the possible branches without silently discarding any that satisfy the original relation.
A single implicit relation may give more than one explicit formula. When that happens, each formula is a branch of the relation. Whether a particular branch is relevant to a differential-equation solution can also depend on an initial condition or the interval under consideration. Here the goal is to solve the relation itself accurately.
Logarithms of Absolute Values
The key equivalence is that, for a real number \(u\neq 0\) and a real number \(v\), \(\ln|u|=v\) exactly when \(|u|=e^v\). Since an absolute value is nonnegative, the equation \(|u|=e^v\) gives two possibilities: \(u=e^v\) or \(u=-e^v\). The exponential \(e^v\) is strictly positive, so neither possibility is zero.
This is an equivalence: each step preserves exactly the solutions of the equation. The \(\pm\) is essential unless the sign of \(u\) is already known from some other condition.
Worked Example: Solving a Logarithmic Relation for \(y\)
Solve \(\ln|y|=2x-1\) for \(y\).
Exponentiate both sides. The right-hand side exponentiates to a positive number:
Removing the absolute value gives two branches:
Equivalently, the two formulas can be written \(y=\pm e^{2x-1}\). To check, either formula gives \(|y|=e^{2x-1}\), so
Both branches satisfy the original relation, and neither has \(y=0\), as required for \(\ln|y|\) to be defined.
When the relation includes an arbitrary integration constant, it is useful to distinguish the positive constant that arises by exponentiating from a signed constant used to combine branches. If \(C\) is real, then \(e^C>0\). Consequently, \(\pm e^C\) can take any nonzero real value, but it cannot equal zero.
where the signed constant \(K\) represents either sign, and its magnitude is \(e^C>0\). This compact notation describes the same nonzero possibilities as writing \(y=\pm A e^{g(x)}\) with \(A=e^C>0\). It does not make \(K=0\) allowable: zero cannot satisfy the original logarithmic relation.
Keeping Track of Absolute Values in \(x\)
Absolute values may involve \(x\) as well as \(y\). In that case, removing them can produce formulas whose appearance depends on whether \(x\) is to the left or right of a point where the inside expression is zero. The original relation still determines the allowed values; splitting into cases is a way to express them without absolute-value notation.
Worked Example: A Logarithmic Relation with \(|x+1|\)
Solve \(\ln|y-2|=\ln|x+1|+\ln 3\) for \(y\).
The logarithms require \(y\neq 2\) and \(x\neq -1\). Combine the right-hand logarithms using \(\ln a+\ln b=\ln(ab)\) for positive \(a,b\):
The quantities inside both logarithms are positive on the stated conditions. Since the natural logarithm is one-to-one, their arguments are equal:
Now take both sign branches:
Thus the two explicit formulas are \(y=2+3|x+1|\) and \(y=2-3|x+1|\), with \(x\neq-1\). To display them without absolute values, consider the two possible signs of \(x+1\).
If \(x>-1\), then \(|x+1|=x+1\), so the branches are \(y=2+3(x+1)=3x+5\) and \(y=2-3(x+1)=-3x-1\).
If \(x<-1\), then \(|x+1|=-(x+1)=-x-1\), so the branches are \(y=2+3(-x-1)=-3x-1\) and \(y=2-3(-x-1)=3x+5\).
For a check, either branch in either case gives \(|y-2|=3|x+1|\). Taking logarithms gives \(\ln|y-2|=\ln(3|x+1|)=\ln|x+1|+\ln 3\), which is the original relation. The point \(x=-1\) is excluded because the original right-hand side contains \(\ln|x+1|\).
Algebraic Relations and Multiple Roots
After exponentiating or rearranging a relation, the next step may be algebraic rather than logarithmic. A familiar source of multiple branches is an equation involving a square. If \((y-a)^2=R(x)\), real values of \(y\) exist only where \(R(x)\geq 0\), and solving gives \(y=a+\sqrt{R(x)}\) or \(y=a-\sqrt{R(x)}\). If \(R(x)=0\), these two formulas give the same value; where \(R(x)>0\), they give distinct values.
Worked Example: Solving a Quadratic Relation for \(y\)
Solve \(y^2-2y=4x+5\) for real \(y\).
Complete the square on the left by adding \(1\) to both sides:
For real \(y\), the right-hand side must be nonnegative. Therefore \(4x+6\geq 0\), or \(x\geq-\frac{3}{2}\). Taking square roots gives both branches:
Check by substitution. For either sign, write \(y=1\pm\sqrt{4x+6}\). Then
Thus both formulas satisfy the original relation wherever the square root is real. At \(x=-\frac{3}{2}\), the two branches meet at \(y=1\); for \(x>-\frac{3}{2}\), they give distinct values.
A Reliable Solving Procedure
The details differ from one relation to another, but the following sequence helps prevent lost branches and invalid values.
Identify where logarithms, denominators, and square roots in the original relation are defined. These restrictions remain in force during rearrangement.
For a logarithm, exponentiate both sides. For a square, take both square-root branches, while requiring a nonnegative radicand for real solutions.
Keep both signs when the equation permits both. Split into cases if an absolute value needs to be removed.
Substitute each proposed branch into the original equation, not just an intermediate rearrangement, and confirm that its restrictions are met.
Why the Branches Matter
The reason for tracking branches is that many operations used to solve an equation are not one-to-one. Squaring can make two different values equal, and an absolute value identifies a number and its negative: \(|u|=|-u|\). As a result, reversing either operation may require more than one possibility.
For example, if one starts from \(|u|=a\) with \(a>0\), the solutions are \(u=a\) and \(u=-a\), not just \(u=a\). Conversely, if a relation already tells us \(u>0\), then only the positive branch is possible. A branch can be removed only when a stated condition rules it out.
This is also why it is important to distinguish an equation from a formula. The equation \((y-1)^2=4x+6\) describes both square-root branches together. Writing \(y=1+\sqrt{4x+6}\) gives only one of them. That may be the branch of interest in a particular application, but it is not the full solution of the relation unless a further condition selects it.
Common Mistake: Dropping the Negative Branch
A frequent error is to exponentiate \(\ln|y|=2x-1\) and then write \(y=e^{2x-1}\). Exponentiating correctly gives \(|y|=e^{2x-1}\), not \(y=e^{2x-1}\). Removing the absolute value requires both signs: \(y=\pm e^{2x-1}\).
The omitted branch is not merely a formal possibility. For \(y=-e^{2x-1}\), the absolute value is \(|y|=e^{2x-1}\), so \(\ln|y|=2x-1\) as well. The negative branch therefore satisfies the original relation exactly. The corrected procedure is to exponentiate first, then solve the absolute-value equation, and finally check each result in the original relation.
Check Your Understanding
For each question, keep track of both the branches and any restrictions inherited from the relation.
- Solve \(\ln|y-4|=x\) explicitly for \(y\).
- Solve \(\ln|y|=\ln|x-2|+1\). Which value of \(x\) is excluded?
- Solve \((y+3)^2=2x-1\) for real \(y\), and state the condition on \(x\).
- Explain why \(y=e^x\) alone does not solve the full relation \(\ln|y|=x\).
- For \(|y+1|=5\), list every value of \(y\) and verify each one in the relation.
