Sr Examen

Otras calculadoras

log(4^x-b,2)=0 la ecuación

El profesor se sorprenderá mucho al ver tu solución correcta😉

v

Solución numérica:

Buscar la solución numérica en el intervalo [, ]

Solución

Ha introducido [src] 
   / x    \    
log\4  - b/    
----------- = 0
   log(2)
log⁡(4x−b)log⁡(2)=0\frac{\log{\left(4^{x} - b \right)}}{\log{\left(2 \right)}} = 0
Simplificar
Gráfica
Respuesta rápida [src] 
        /|  _______|\        /   _______\
     log\|\/ 1 + b |/   I*arg\-\/ 1 + b /
x1 = ---------------- + -----------------
          log(2)              log(2)
x1=log⁡(∣b+1∣)log⁡(2)+iarg⁡(−b+1)log⁡(2)x_{1} = \frac{\log{\left(\left|{\sqrt{b + 1}}\right| \right)}}{\log{\left(2 \right)}} + \frac{i \arg{\left(- \sqrt{b + 1} \right)}}{\log{\left(2 \right)}} 
     log(|1 + b|)   I*arg(1 + b)
x2 = ------------ + ------------
       2*log(2)       2*log(2)
x2=log⁡(∣b+1∣)2log⁡(2)+iarg⁡(b+1)2log⁡(2)x_{2} = \frac{\log{\left(\left|{b + 1}\right| \right)}}{2 \log{\left(2 \right)}} + \frac{i \arg{\left(b + 1 \right)}}{2 \log{\left(2 \right)}} 
x2 = log(|b + 1|)/(2*log(2)) + i*arg(b + 1)/(2*log(2))
Suma y producto de raíces [src] 
suma
   /|  _______|\        /   _______\                              
log\|\/ 1 + b |/   I*arg\-\/ 1 + b /   log(|1 + b|)   I*arg(1 + b)
---------------- + ----------------- + ------------ + ------------
     log(2)              log(2)          2*log(2)       2*log(2)
(log⁡(∣b+1∣)log⁡(2)+iarg⁡(−b+1)log⁡(2))+(log⁡(∣b+1∣)2log⁡(2)+iarg⁡(b+1)2log⁡(2))\left(\frac{\log{\left(\left|{\sqrt{b + 1}}\right| \right)}}{\log{\left(2 \right)}} + \frac{i \arg{\left(- \sqrt{b + 1} \right)}}{\log{\left(2 \right)}}\right) + \left(\frac{\log{\left(\left|{b + 1}\right| \right)}}{2 \log{\left(2 \right)}} + \frac{i \arg{\left(b + 1 \right)}}{2 \log{\left(2 \right)}}\right) 
=
   /|  _______|\                       /   _______\               
log\|\/ 1 + b |/   log(|1 + b|)   I*arg\-\/ 1 + b /   I*arg(1 + b)
---------------- + ------------ + ----------------- + ------------
     log(2)          2*log(2)           log(2)          2*log(2)
log⁡(∣b+1∣)log⁡(2)+log⁡(∣b+1∣)2log⁡(2)+iarg⁡(−b+1)log⁡(2)+iarg⁡(b+1)2log⁡(2)\frac{\log{\left(\left|{\sqrt{b + 1}}\right| \right)}}{\log{\left(2 \right)}} + \frac{\log{\left(\left|{b + 1}\right| \right)}}{2 \log{\left(2 \right)}} + \frac{i \arg{\left(- \sqrt{b + 1} \right)}}{\log{\left(2 \right)}} + \frac{i \arg{\left(b + 1 \right)}}{2 \log{\left(2 \right)}} 
producto
/   /|  _______|\        /   _______\\                              
|log\|\/ 1 + b |/   I*arg\-\/ 1 + b /| /log(|1 + b|)   I*arg(1 + b)\
|---------------- + -----------------|*|------------ + ------------|
\     log(2)              log(2)     / \  2*log(2)       2*log(2)  /
(log⁡(∣b+1∣)log⁡(2)+iarg⁡(−b+1)log⁡(2))(log⁡(∣b+1∣)2log⁡(2)+iarg⁡(b+1)2log⁡(2))\left(\frac{\log{\left(\left|{\sqrt{b + 1}}\right| \right)}}{\log{\left(2 \right)}} + \frac{i \arg{\left(- \sqrt{b + 1} \right)}}{\log{\left(2 \right)}}\right) \left(\frac{\log{\left(\left|{b + 1}\right| \right)}}{2 \log{\left(2 \right)}} + \frac{i \arg{\left(b + 1 \right)}}{2 \log{\left(2 \right)}}\right) 
=
/     /   _______\      /|  _______|\\                              
\I*arg\-\/ 1 + b / + log\|\/ 1 + b |//*(I*arg(1 + b) + log(|1 + b|))
--------------------------------------------------------------------
                                  2                                 
                             2*log (2)
(log⁡(∣b+1∣)+iarg⁡(−b+1))(log⁡(∣b+1∣)+iarg⁡(b+1))2log⁡(2)2\frac{\left(\log{\left(\left|{\sqrt{b + 1}}\right| \right)} + i \arg{\left(- \sqrt{b + 1} \right)}\right) \left(\log{\left(\left|{b + 1}\right| \right)} + i \arg{\left(b + 1 \right)}\right)}{2 \log{\left(2 \right)}^{2}} 
(i*arg(-sqrt(1 + b)) + log(Abs(sqrt(1 + b))))*(i*arg(1 + b) + log(|1 + b|))/(2*log(2)^2)
    © Sr Examen – Калькулятор