- #1

murshid_islam

- 442

- 17

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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