# How do you solve  log(x-5) + log(x-2)=1?

Nov 29, 2015

$x = 7$

#### Explanation:

$\log \left(x - 5\right) + \log \left(x - 2\right) = 1$
$\textcolor{w h i t e}{\text{XXXXXXXX}}$since $\log \left(a \cdot b\right) = \log \left(a\right) + \log \left(b\right)$
$\Rightarrow \log \left(\left(x - 5\right) \left(x - 2\right)\right) = 1$
$\textcolor{w h i t e}{\text{XXXXXXXX}}$expanding the multiplication
$\log \left({x}^{2} - 7 x + 10\right) = 1$
$\textcolor{w h i t e}{\text{XXXXXXXX}}$despite other comments the default base for $\log$ is $10$
$\textcolor{w h i t e}{\text{XXXXXXXX}}$so using the above as exponents of $10$
${10}^{\log \left(2 {x}^{2} - 7 x + 10\right)} = {10}^{1}$
$\Rightarrow {x}^{2} - 7 x + 10 = 10$
$\textcolor{w h i t e}{\text{XXXXXXXX}}$subtracting $10$ from both sides
${x}^{2} - 7 x = 0$
$\textcolor{w h i t e}{\text{XXXXXXXX}}$factoring
$x \left(x - 7\right) = 0$

i.e. $x = 0$ or $x = 7$

However $\log \left(x - 5\right)$ and $\log \left(x - 2\right)$ are undefined if $x = 0$;
so $x = 0$ is an extraneous solution
and only $x = 7$ is valid

Sep 3, 2017

$x = 7$

#### Explanation:

As no base is given, it is assumed to be $10$
Natural logs are commonly denoted by ln.

$\log \left(x - 5\right) + \log \left(x - 2\right) = \textcolor{b l u e}{1}$

In an expression or equation, the terms must all be in the same form - either all logs or all numbers.

$\log \left(x - 5\right) + \log \left(x - 2\right) = \textcolor{b l u e}{\log 10} \text{ } \leftarrow \left(\textcolor{b l u e}{{\log}_{10} 10 \Leftrightarrow 1}\right)$

$\textcolor{w h i t e}{\times \times \times \times \times x}$Apply the law: $\text{ } \log a + \log b \Leftrightarrow \log \left(a b\right)$

$\log \left(\left(x - 5\right) \left(x - 2\right)\right) = \log 10$

$\textcolor{w h i t e}{\times \times \times \times \times x}$Apply the law: $\text{ } \log a = \log b \Leftrightarrow a = b$

$\therefore \left(x - 5\right) \left(x - 2\right) = 10 \textcolor{w h i t e}{\times \times \times \times \times x}$'drop' the logs

${x}^{2} - 7 x + 10 = 10 \text{ } \leftarrow$ solve the quadratic equation

${x}^{2} - 7 x = 0 \text{ } \leftarrow$ factorise

$x \left(x - 7\right) = 0$

$x = 0 \mathmr{and} x = 7$

However, $x = 0$ is an extraneous solution, and not valid in this equation.

$x = 7$