# Transcendental equation

In applied mathematics, a transcendental equation is an equation over the real (or complex) numbers that is not algebraic, that is, if at least one of its sides describes a transcendental function. Examples include:

{\begin{aligned}x&=e^{-x}\\x&=\cos x\\2^{x}&=x^{2}\end{aligned}} 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:

$4^{x}=3^{x^{2}-1}\cdot 2^{5x}$ transforms to $x\ln 4=(x^{2}-1)\ln 3+5x\ln 2$ , which simplifies to $x^{2}\ln 3+x(5\ln 2-\ln 4)-\ln 3=0$ , which has the solutions $x={\frac {-3\ln 2\pm {\sqrt {9(\ln 2)^{2}-4(\ln 3)^{2}}}}{2\ln 3}}.$ This will not work if addition occurs "at the base line", as in $4^{x}=3^{x^{2}-1}+2^{5x}.$ • if all "base constants" can be written as integer or rational powers of some number q, then substituting y=qx may succeed, e.g.
$2^{x-1}+4^{x-2}-8^{x-2}=0$ transforms, using y=2x, to ${\frac {1}{2}}y+{\frac {1}{16}}y^{2}-{\frac {1}{64}}y^{3}=0$ which has the solutions $y\in \{0,-4,8\}$ , hence $x=\log _{2}8=3$ 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=xex may obtain an algebraic equation; after the solutions for y are known, those for x can be obtained by applying the Lambert W function,[citation needed] e.g.:
$x^{2}e^{2x}+2=3xe^{x}$ transforms to $y^{2}+2=3y,$ which has the solutions $y\in \{1,2\},$ hence $x\in \{W_{0}(1),W_{0}(2),W_{-1}(1),W_{-1}(2)\}$ , where $W_{0}$ and $W_{-1}$ the denote the real-valued branches of the multivalued $W$ 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.
$2\log _{5}(3x-1)-\log _{5}(12x+1)=0$ transforms, using exponentiation to base $5.$ to ${\frac {(3x-1)^{2}}{12x+1}}=1,$ which has the solutions $x\in \{0,2\}.$ If only real numbers are considered, $x=0$ is not a solution, as it leads to a non-real subexpression $\log _{5}(-1)$ 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 $b$ and a unique argument expression $f(x),$ then substituting $y=\log _{b}(f(x))$ may lead to a simpler equation, e.g.
$5\ln(\sin x^{2})+6=7{\sqrt {\ln(\sin x^{2})+8}}$ transforms, using $y=\ln(\sin x^{2}),$ to $5y+6=7{\sqrt {y+8}},$ which is algebraic and can be solved.[clarification needed] After that, applying inverse operations to the substitution equation yields ${\sqrt {\arcsin \exp y}}=x.$ ### 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 $\sin(nx+a),\cos(mx+b),\tan(lx+c),...$ with integer $n,m,l,...$ might all be transformed to arguments of the form, say, $\sin x$ . After that, substituting $y=\sin(x)$ yields an algebraic equation, e.g.
$\sin(x+a)=(\cos ^{2}x)-1$ transforms to $(\sin x)(\cos a)+{\sqrt {1-\sin ^{2}x}}(\sin a)=1-(\sin ^{2}x)-1$ , and, after substitution, to $y(\cos a)+{\sqrt {1-y^{2}}}(\sin a)=-y^{2}$ which is algebraic and can be solved. After that, applying $x=2k\pi +\arcsin y$ 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 $y=exp(x)$ yields an algebraic equation, e.g.
$3\cosh x=4+\sinh(2x-6)$ unfolds to ${\frac {3}{2}}(e^{x}+{\frac {1}{e^{x}}})=4+{\frac {1}{2}}\left({\frac {(e^{x})^{2}}{e^{6}}}-{\frac {e^{6}}{(e^{x})^{2}}}\right),$ which transforms to the equation ${\frac {3}{2}}(y+{\frac {1}{y}})=4+{\frac {1}{2}}\left({\frac {y^{2}}{e^{6}}}-{\frac {e^{6}}{y^{2}}}\right),$ which is algebraic and can be solved. Applying $x=\ln y$ obtains the solutions of the original equation.

## Approximate solutions

Approximate numerical solutions to transcendental equations can be found using numerical, analytical approximations, or graphical methods.

Numerical methods for solving arbitrary equations are called root-finding algorithms.

In some cases, the equation can be well approximated using Taylor series near the zero. For example, for $k\approx 1$ , the solutions of $\sin x=kx$ are approximately those of $(1-k)x-x^{3}/6=0$ , namely $x=0$ and $x=\pm {\sqrt {6}}{\sqrt {1-k}}$ .

For a graphical solution, one method is to set each side of a single-variable transcendental equation equal to a dependent variable and plot the two graphs, using their intersecting points to find solutions (see picture).

## Other solutions

• Some transcendental systems of high-order equations can be solved by “separation” of the unknowns, reducing them to algebraic equations.
• The following can also be used when solving transcendental equations/inequalities: If $x_{0}$ is a solution to the equation $f(x)=g(x)$ and $f(x)\leq c\leq g(x)$ , then this solution must satisfy $f(x_{0})=g(x_{0})=c$ . For example, we want to solve $\log _{2}\left(3+2x-x^{2}\right)=\tan ^{2}\left({\frac {\pi x}{4}}\right)+\cot ^{2}\left({\frac {\pi x}{4}}\right)$ . The given equation is defined for $-1 . Let $f(x)=\log _{2}\left(3+2x-x^{2}\right)$ and $g(x)=\tan ^{2}\left({\frac {\pi x}{4}}\right)+\cot ^{2}\left({\frac {\pi x}{4}}\right)$ . It is easy to show that $f(x)\leq 2$ and $g(x)\geq 2$ so if there is a solution to the equation, it must satisfy $f(x)=g(x)=2$ . From $f(x)=2$ we get $x=1\in (-1,3)$ . Indeed, $f(1)=g(1)=2$ and so $x=1$ is the only real solution to the equation.