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

Dec 6, 2015

$x = 5$

#### Explanation:

Rearrange the expression to get

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

Now, use the property $\log \left(a\right) + \log \left(b\right) = \log \left(a \cdot b\right)$:

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

If by $\log \left(x\right)$ you mean the logarithm in base $10$, you can write $1$ as $\log \left(10\right)$, and so the expression becomes

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

Now use the fact that $\log \left(a\right) = \log \left(b\right) \setminus \iff a = b$:

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

Expand:

${x}^{2} - 3 x - 10 = 0$

To solve this equation, we need two numbers ${x}_{1}$ and ${x}_{2}$ such that:

${x}_{1} + {x}_{2} = 3$
${x}_{1} \cdot {x}_{2} = - 10$

These numbers are $5$ and $- 2$. We can't accept $- 2$ as a solution, because it would lead to the logarithm of a negative number.