Question

4^×-1 + 2^×+2=48 (Ecuaciones exponenciales . ) Porfis​

Answer 1

Respuesta:

$x=\frac{\ln \left(\frac{-1+3\sqrt{21}}{2}\right)}{\ln \left(2\right)}$

Explicación paso a paso:

$4^x-1+2^x+2=48\\\\\left(2^x\right)^2-1+2^x+2=48\\\\\left(u\right)^2-1+u+2=48\\\\u=\frac{-1+3\sqrt{21}}{2},\:u=\frac{-1-3\sqrt{21}}{2}\\\\x=\frac{\ln \left(\frac{-1+3\sqrt{21}}{2}\right)}{\ln \left(2\right)}$

Answer 2

4× -1 + 2× + 2= 48

4× + 2× = 48 -2 + 1

4× + 2× = 47

4× = √(47 + 2)

4× = √49

4× = 7

× = √(4+7)

×= √11