Solve the following equation: log (x-1) +logx=1?

Question

please help!

Answer 1

log x(x-1)=1

x(x-1)=10

x^2-x-10=0

x= (1+-sqrt(41))/2

DON’T FORGET TO CHECK DOMAIN

x>0

x-1>0 => x>1      D=(1,+infinity)

so  x= (1-sqrt(41))/2 is NOT acceptable

Answer 2

A the sum of logs is the log of the products.  So:

log(x-1)+log(x)=1

log[(x-1)(x)]=1

log(x^2-x)=1    assuming common log (base 10)

x^2-x=10

x^2-x-10=0

Use the quadratic solution to finish this; the answer is irrational.

Answer 3

Assuming base 10,

log(x^2+x)=1

x^2+x=10

x^2+x-10=0

not factorable over the reals

Answer 4

log((x-1)(x)) = 1 { property of logarithms that addition of two logarithms is equal to product of the subject, and if they are subtracting, you just divide them)

Therfore,

log (x^2 - x ) = 1

10^1 = x^2 - x { log is to base 10 unless otherwise specified).

x^2 - x - 10 = 0

Use the quadratic formula, to solve for x: I would write all that but it is too much writing, and i m sure you know quadratic formula.

a = 1

b = -1

c = -10

and you this should be it.. good luck!

*edit*

Natural log and Log are two different things... natural log is to the base "e" which is 2.718....

and log can be to base anything but if no base is written under LOG, it is assumed to be base 10.

you can not write LOG or LN interchangeably.

Answer 5

I assume this is base-10 log?  (Natural log is usually marked as LN, but some people mark it as LOG.)

log (x - 1) + log x = 1

log [(x - 1)(x)] = 1

(x - 1)(x) = 10^1

x^2 - x = 10

x^2 - x - 10 = 0

Use quadratic equation to finish solving for x.