# How do you solve log(x-3)+log x=1?

Apr 28, 2018

5

#### Explanation:

log(x−3)+logx=1

logarithms on the left side may be added together by multiplying argumnts, on the other side we can rewrite number 1

log[(x−3)*x]=log10

Logarithm is a simple function, therefore we can compare arguments

[(x−3)*x]=10

x^2−3x-10=0

$\left(x + 2\right) \cdot \left(x - 5\right) = 0$

${x}_{1} = - 2$
${x}_{2} = 5$

Since argument of logaritm can be only positive, ${x}_{1}$ is not a solution
log(-2−3)+log(-2)=> not possible

log(5−3)+log5=>log2+log5=>log(5*2)=>log10=>1
$1 = 1$ ;correct;

Apr 28, 2018

See below

#### Explanation:

The goal with type of problems is to get a expresion like $\log A = \log B$. By injectivity of function log, we can say that $A = B$

Let see...
Using logarithmic rules

$\log \left(x - 3\right) + \log x = \log 10 = 1$

$\log \left(x - 3\right) x = \log 10$

Then $\left(x - 3\right) x = 10$

However 10=5x2 or 2x5, or negatives ones. Let say $x - 3 = 5$ then $x = 2$

If $x = 5$ then $x - 3 = 2$

Other way is operate in $x \left(x - 3\right) = {x}^{2} - 3 x = 10$ or

${x}^{2} - 3 x - 10 = 0$ using quadratic formula we have

$x = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm 7}{2}$ this arrives to $x = 5$ and $x = - 2$

Lets check the answers $\log \left(- 2 - 3\right) + \log - 2 = 1$ reject this solution because nor $\log - 5$ neither $\log - 2$ doesn`t exists

By other hand $\log \left(5 - 3\right) + \log 5 = \log \left(2 \times 5\right) = \log 10 = 1$

So, the only solution is $x = 5$