Transcendental equation

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

• applying the natural logarithm to both sides may yield an algebraic equation,[3] 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.[4]

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.:

${\displaystyle x^{2}e^{2x}+2=3xe^{x}}$ 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 ${\displaystyle W_{-1}}$ the denote the real-valued branches of the multivalued ${\displaystyle W}$ function.

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.[5]

• 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,[6] 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 ${\displaystyle y={\frac {89}{25}}}$.[lower-alpha 2] After that, applying inverse operations to the substitution equation yields ${\displaystyle x={\sqrt {\arcsin \exp y}}={\sqrt {\arcsin \exp {\frac {89}{25}}}}.}$

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,[7] 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[lower-alpha 3] and can be solved. After that, applying ${\displaystyle x=2k\pi +\arcsin y}$ obtains the solutions.

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,[8] 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[lower-alpha 4] and can be solved. Applying ${\displaystyle x=\ln y}$ obtains the solutions of the original equation.