25^x+5^(2*x-1)=150 la ecuación
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Solución
Solución detallada
Tenemos la ecuación:
$$25^{x} + 5^{2 x - 1} = 150$$
o
$$\left(25^{x} + 5^{2 x - 1}\right) - 150 = 0$$
Sustituimos
$$v = 25^{x}$$
obtendremos
$$5^{2 x - 1} + v - 150 = 0$$
o
$$5^{2 x - 1} + v - 150 = 0$$
hacemos cambio inverso
$$25^{x} = v$$
o
$$x = \frac{\log{\left(v \right)}}{\log{\left(25 \right)}}$$
Entonces la respuesta definitiva es
$$x_{1} = \frac{\log{\left(\frac{3}{2} \right)}}{\log{\left(25 \right)}} = \log{\left(\left(\frac{3}{2}\right)^{\frac{1}{\log{\left(25 \right)}}} \right)}$$
$$x_{2} = \frac{\log{\left(\frac{\frac{\log{\left(125 \right)}}{2} + i \pi}{\log{\left(5 \right)}} \right)}}{\log{\left(25 \right)}} = \frac{\log{\left(\frac{\log{\left(125 \right)} + 2 i \pi}{2 \log{\left(5 \right)}} \right)}}{\log{\left(25 \right)}}$$
Suma y producto de raíces
[src]
3 log(125) pi*I
- + -------- + ------
2 2*log(5) log(5)
$$\frac{3}{2} + \left(\frac{\log{\left(125 \right)}}{2 \log{\left(5 \right)}} + \frac{i \pi}{\log{\left(5 \right)}}\right)$$
3 log(125) pi*I
- + -------- + ------
2 2*log(5) log(5)
$$\frac{3}{2} + \frac{\log{\left(125 \right)}}{2 \log{\left(5 \right)}} + \frac{i \pi}{\log{\left(5 \right)}}$$
/log(125) pi*I \
3*|-------- + ------|
\2*log(5) log(5)/
---------------------
2
$$\frac{3 \left(\frac{\log{\left(125 \right)}}{2 \log{\left(5 \right)}} + \frac{i \pi}{\log{\left(5 \right)}}\right)}{2}$$
9 3*pi*I
- + --------
4 2*log(5)
$$\frac{9}{4} + \frac{3 i \pi}{2 \log{\left(5 \right)}}$$
$$x_{1} = \frac{3}{2}$$
log(125) pi*I
x2 = -------- + ------
2*log(5) log(5)
$$x_{2} = \frac{\log{\left(125 \right)}}{2 \log{\left(5 \right)}} + \frac{i \pi}{\log{\left(5 \right)}}$$
x2 = log(125)/(2*log(5)) + i*pi/log(5)
x2 = 1.5 + 1.95198126583117*i
x2 = 1.5 + 1.95198126583117*i