- #1
murshid_islam
- 457
- 19
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] .
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