Transcendental equation

${\displaystyle \begin{array}{cc}x& =e^{-x}\\ x& =\cos x\\ 2^x& =x^2\end{array}}$

A transcendental equation need not be an equation between elementary functions, although most published examples are.

In some cases, a transcendental equation can be solved by transforming it into an equivalent algebraic equation. Some such transformations are sketched below; computer algebra systems may provide more elaborated transformations.

In general, however, only approximate solutions can be found.

Transformation into an algebraic equation

Ad hoc methods exist for some classes of transcendental equations in one variable to transform them into algebraic equations which then might be solved.

Exponential equations

If the unknown, say x, occurs only in exponents:

• applying the natural logarithm to both sides may yield an algebraic equation, e.g.

${\displaystyle 4^x=3^{x^2-1}\cdot 2^{5x}}$ transforms to ${\displaystyle x\ln 4=(x^2-1)\ln 3+5x\ln 2}$, which simplifies to ${\displaystyle x^2\ln 3+x(5\ln 2-\ln 4)-\ln 3=0}$, which has the solutions ${\displaystyle x=\frac{-3\ln 2\pm \sqrt{9(\ln 2{)}^2-4(\ln 3{)}^2}}{2\ln 3}.}$

This will not work if addition occurs "at the base line", as in ${\displaystyle 4^x=3^{x^2-1}+2^{5x}.}$

• if all "base constants" can be written as integer or rational powers of some number q, then substituting y=q^x may succeed, e.g.

${\displaystyle 2^{x-1}+4^{x-2}-8^{x-2}=0}$ transforms, using y=2^x, to ${\displaystyle \frac{1}{2}y+\frac{1}{16}{y}^2-\frac{1}{64}{y}^3=0}$ which has the solutions ${\displaystyle y\in \{0,-4,8\}}$, hence ${\displaystyle x={\log }_{2}8=3}$ is the only real solution.

This will not work if squares or higher power of x occurs in an exponent, or if the "base constants" do not "share" a common q.

• sometimes, substituting y=xe^x may obtain an algebraic equation; after the solutions for y are known, those for x can be obtained by applying the Lambert W function, e.g.:

${\displaystyle x^2{e}^{2x}+2=3x{e}^x}$ transforms to ${\displaystyle y^2+2=3y,}$ which has the solutions ${\displaystyle y\in \{1,2\},}$ hence ${\displaystyle x\in \{W_0(1),W_0(2),W_{-1}(1),W_{-1}(2)\}}$, where ${\displaystyle W_0}$ and ${\displaystyle W_{-1}}$ denote the real-valued branches of the multivalued ${\displaystyle W}$ function.

Logarithmic equations

If the unknown x occurs only in arguments of a logarithm function:

• applying exponentiation to both sides may yield an algebraic equation, e.g.

${\displaystyle 2{\log }_5(3x-1)-{\log }_5(12x+1)=0}$ transforms, using exponentiation to base ${\displaystyle 5.}$ to ${\displaystyle \frac{(3x-1{)}^2}{12x+1}=1,}$ which has the solutions ${\displaystyle x\in \{0,2\}.}$ If only real numbers are considered, ${\displaystyle x=0}$ is not a solution, as it leads to a non-real subexpression ${\displaystyle {\log }_5(-1)}$ in the given equation.

This requires the original equation to consist of integer-coefficient linear combinations of logarithms w.r.t. a unique base, and the logarithm arguments to be polynomials in x.

• if all "logarithm calls" have a unique base ${\displaystyle b}$ and a unique argument expression ${\displaystyle f(x),}$ then substituting ${\displaystyle y={\log }_b(f(x))}$ may lead to a simpler equation, e.g.

${\displaystyle 5\ln (\sin x^2)+6=7\sqrt{\ln (\sin x^2)+8}}$ transforms, using ${\displaystyle y=\ln (\sin x^2),}$ to ${\displaystyle 5y+6=7\sqrt{y+8},}$ which is algebraic and has the single solution ${\displaystyle y=\frac{89}{25}}$. After that, applying inverse operations to the substitution equation yields ${\displaystyle x=\sqrt{\arcsin \exp y}=\sqrt{\arcsin \exp \frac{89}{25}}.}$

Trigonometric equations

If the unknown x occurs only as argument of trigonometric functions:

• applying Pythagorean identities and trigonometric sum and multiple formulas, arguments of the forms ${\displaystyle \sin (nx+a),\cos (mx+b),\tan (lx+c),...}$ with integer ${\displaystyle n,m,l,...}$ might all be transformed to arguments of the form, say, ${\displaystyle \sin x}$. After that, substituting ${\displaystyle y=\sin (x)}$ yields an algebraic equation, e.g.

${\displaystyle \sin (x+a)=({\cos }^{2}x)-1}$ transforms to ${\displaystyle (\sin x)(\cos a)+\sqrt{1-{\sin }^{2}x}(\sin a)=1-({\sin }^{2}x)-1}$, and, after substitution, to ${\displaystyle y(\cos a)+\sqrt{1-y^2}(\sin a)=-y^2}$ which is algebraic and can be solved. After that, applying ${\displaystyle x=2k\pi +\arcsin y}$ obtains the solutions.

Hyperbolic equations

If the unknown x occurs only in linear expressions inside arguments of hyperbolic functions,

• unfolding them by their defining exponential expressions and substituting ${\displaystyle y=\exp (x)}$ yields an algebraic equation, e.g.

${\displaystyle 3\cosh x=4+\sinh (2x-6)}$ unfolds to ${\displaystyle \frac{3}{2}(e^x+\frac{1}{e^x})=4+\frac{1}{2}\left(\frac{(e^x{)}^2}{e^6}-\frac{e^6}{(e^x{)}^2}\right),}$ which transforms to the equation ${\displaystyle \frac{3}{2}(y+\frac{1}{y})=4+\frac{1}{2}\left(\frac{y^2}{e^6}-\frac{e^6}{y^2}\right),}$ which is algebraic and can be solved. Applying ${\displaystyle x=\ln y}$ obtains the solutions of the original equation.