How do you solve #\ln ( x - 1) = 5#?

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Answer 1 · 2018-03-10T00:37:02

The exact solution is #x=e^5+1#.

Rewrite the equation in exponential form (as opposed to #log# form):

#log_color(red)acolor(green)y=color(blue)xqquad<=>qquadcolor(red)a^color(blue)x=color(green)y#

Here's the actual problem:

#color(white)=>ln(x-1)=5#

#color(white)=>log_color(red)ecolor(green)((x-1))=color(blue)5#

#=>color(red)e^color(blue)5=color(green)(x-1)#

#color(white)=>color(red)e^color(blue)5color(green)+color(green)1=color(green)(x)#

#color(white)=>x=e^5+1#

#color(white)(=>x)~~149.413...#