Question

what is log(x-3) + logx=1

Answer 1

\(\large\color{black}{ \log(a)+\log(b) =\log(a\times b) }\)

Answer 2

your "a" is x-3
and
your "b" is x.

Answer 3

x = 5

Answer 4

why ???? \(\color{green}{why}\)

Answer 5

This may help:

http://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.272775.html

Answer 6

No fighting with me please.

Answer 7

\(\large\color{black}{ \log(a)+\log(b) =\log(a\times b) }\)
when you re-write it in your problem, it comes out the same way.

\(\large\color{black}{ \log(x-3)+\log(x) =\log([x-3]\times x) =\log(x^2-3x) }\)

Answer 8

no @OpiGeode we are not fighting with you
just try to tell you that direct answer not gonna help student
:) keep smiling :)

Answer 9

And since you know that
log(x-3) + log(x) = 1
you can say as your next step,

\(\large\color{black}{ \log(x^2-3x)=1 }\)

Answer 10

When the base is unspecified it is 10.

Answer 11

wait for what ?

Answer 12

The direct answer has been given away, so I want to post the solution.

Answer 13

ok :) :) :)

Answer 14

\(\large\color{black}{ \log_{10}(x^2-3x)=1 }\)
there is a rule,
\(\large\color{black}{ \log_{a}(b)=c~~~~~->~~~~~(a)^c=b }\)
So,
\(\large\color{black}{ \log_{10}(x^2-3x)=1~~~~~->~~~~~(x^2-3x)^1=10 }\)
And, then....
\(\large\color{black}{ (x^2-3x)^1=10 }\)
\(\large\color{black}{ x^2-3x=10 }\)
\(\large\color{black}{ x^2-3x-10=0 }\)

comes down to a quadratic function.

Answer 15

Factoring would be the fastest.
\(\large\color{black}{ x^2-3x-10=0 }\)
\(\large\color{black}{ (x-5)(x+2)=0 }\)
\(\large\color{black}{ x=-2,~~5 }\)

Answer 16

Although the quadratic has 2 soltuions, the only one that works is 5.

Answer 17

because the initial question is
\(\large\color{black}{ \log(x-3) + \log(x)=1 }\)
when you have x=-2, you get
\(\large\color{black}{ \log(-5) + \log(-2)=1 }\)

BUT log of a number that is zero or less than zero, is undefined.