# How do you solve Logx + Log(x+9)=1?

May 16, 2018

See below

#### Explanation:

Our goal is to have an expresion like this $\log A = \log B$ due to one to one aplication of logarithm, we get $A = B$. We have to apply the rules of logarithm for product, quotient and power and use the fact $\log 10 = 1$

$\log x + \log \left(x + 9\right) = 1$
$\log x \left(x + 9\right) = \log 10$

Now, we know that $x \left(x + 9\right) = 10$
${x}^{2} + 9 x - 10 = 0$ using quadratic formula we get

$x = - 10 \mathmr{and} x = 1$ reject negative because there is no logarithm and the only solution is $x = 1$. Lets check it

$\log 1 + \log \left(1 + 9\right) = 0 + 1 = 1$

May 16, 2018

$x = 1$

#### Explanation:

$\log x + \log \left(x + 9\right) = 1$

$\implies \log \left(x \left(x + 9\right)\right) = 1$

i.e. $x \left(x + 9\right) = 10$

or ${x}^{2} + 9 x - 10 = 0$

or ${x}^{2} - x + 10 x - 10 = 0$

or $x \left(x - 1\right) + 10 \left(x - 1\right) = 0$

or $\left(x - 1\right) \left(x + 10\right) = 0$

so either $x - 1 = 0$ i.e. $x = 1$

or $x + 10 = 0$ i.e. $x = - 10$

but we cannot have $x \le - 9$ as $x + 9 > 0$

Hence only solution is $x = 1$