Resolver {l}{2^{3x-1}*8^{x+2}=32^{4-1x}}{(3^x+4)^x-2=5^x^2-x-5}

Ebatzi: x

x=\frac{2\pi n_{1}i}{11\ln(2)}+\frac{15}{11}

n_{1}\in \mathrm{Z}

\left(81\times 3^{\frac{2\pi n_{1}i}{11\ln(2)}+\frac{15}{11}}\right)^{\frac{2\pi n_{1}i}{11\ln(2)}-\frac{7}{11}}-5^{\left(\frac{2\pi n_{1}i}{11\ln(2)}+\frac{15}{11}\right)^{2}-\frac{2\pi n_{1}i}{11\ln(2)}-\frac{70}{11}}=0

Azterketa

Complex Number

\left. \begin{array} { l } { 2 ^ { 3 x - 1 } \cdot 8 ^ { x + 2 } = 32 ^ { 4 - 1 x } } \\ { ( 3 ^ { x + 4 } ) ^ { x - 2 } = 5 ^ { x ^ { 2 } - x - 5 } } \end{array} \right.

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