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:
x = e − x x = cos x 2 x = x 2 {\displaystyle {\begin{aligned}x&=e^{-x}\\x&=\cos x\\2^{x}&=x^{2}\end{aligned}}}
A transcendental equation may involve also non-elementary functions, although most published examples do not. 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.
4 x = 3 x 2 − 1 ⋅ 2 5 x {\displaystyle 4^{x}=3^{x^{2}-1}\cdot 2^{5x}} transforms to x ln 4 = ( x 2 − 1 ) ln 3 + 5 x ln 2 {\displaystyle 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 {\displaystyle x^{2}\ln 3+x(5\ln 2-\ln 4)-\ln 3=0} , which has the solutions x = − 3 ln 2 ± 9 ( ln 2 ) 2 − 4 ( ln 3 ) 2 2 ln 3 . {\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 4 x = 3 x 2 − 1 + 2 5 x . {\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=qx may succeed, e.g.
2 x − 1 + 4 x − 2 − 8 x − 2 = 0 {\displaystyle 2^{x-1}+4^{x-2}-8^{x-2}=0} transforms, using y=2x, to 1 2 y + 1 16 y 2 − 1 64 y 3 = 0 {\displaystyle {\frac {1}{2}}y+{\frac {1}{16}}y^{2}-{\frac {1}{64}}y^{3}=0} which has the solutions y ∈ { 0 , − 4 , 8 } {\displaystyle y\in \{0,-4,8\}} , hence x = log 2 8 = 3 {\displaystyle x=\log _{2}8=3} is the only real solution. This will not work if a square or a 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, e.g.:
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