- #1

- 388

- 3

i have to solve for x:

[tex]x^a - x = 1[/tex] where [tex]a = \frac{\ln 6}{\ln 5}[/tex]

taking [itex]\ln[/itex] in both side, i get,

[tex]a\ln x = \ln(x+1)[/tex]

[tex]\Rightarrow\frac{\ln(x+1)}{\ln x} = a = \frac{\ln 6}{\ln 5}[/tex]

here we can see that x = 5.

but what i wanted to know is the general solution of the equation for any [itex]a \in \mathbb{R}[/itex] .

[tex]x^a - x = 1[/tex] where [tex]a = \frac{\ln 6}{\ln 5}[/tex]

taking [itex]\ln[/itex] in both side, i get,

[tex]a\ln x = \ln(x+1)[/tex]

[tex]\Rightarrow\frac{\ln(x+1)}{\ln x} = a = \frac{\ln 6}{\ln 5}[/tex]

here we can see that x = 5.

but what i wanted to know is the general solution of the equation for any [itex]a \in \mathbb{R}[/itex] .

Last edited: