# How do you solve log (x+9) - log x = 1?

Dec 14, 2015

$x = 1$

#### Explanation:

1) Determine when the equation is defined

First of all, let's determine for which $x$ your equation is defined and for which $x$ it is not defined.

Any logarithmic expression is only defined if its argument is greater than $0$.

So, in your case, $x + 9 > 0 \iff x > - 9$ and $x > 0$ must hold.

As $x > 0$ is the more restrictive condition of the two, the domain of your function is $x > 0$.

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2) Simplify the logarithmic equation

As next, you should combine all your logarithmic expressions into one.

You can do this with the logarithmic law

${\log}_{a} \left(n\right) - {\log}_{a} \left(m\right) = {\log}_{a} \frac{n}{\log} _ a \left(m\right)$

Thus, you equation can be transformed as follows:

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

$\iff \log \left(\frac{x + 9}{x}\right) = 1$

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3) Eliminate the logarithmic term

Now, as you haven't specified the base of the logarithm, I will assume that the base is $10$.

The inverse function of ${\log}_{10} \left(x\right)$ is ${10}^{x}$ which means that both
${\log}_{10} \left({10}^{x}\right) = x$ and ${10}^{{\log}_{10} \left(x\right)} = x$ hold.

Thus, to eliminate the logarithmic expression at the left side, you need to apply the function ${10}^{x}$ to the both sides of the equation:

$\iff {10}^{{\log}_{10} \left(\frac{x + 9}{x}\right)} = {10}^{1}$

$\iff \frac{x + 9}{x} = 10$

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4) Solve the equation

$\frac{x + 9}{x} = 10$

... multiply both sides with $x$ ...

$\iff x + 9 = 10 x$

$\iff 9 = 9 x$

$\iff x = 1$

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5) Check if the result is valid

We need to check if our result $x = 1$ is consistent with the condition that we have computed in the beginning, $x > 0$.

Here, this is the case, thus we can accept $x = 1$ as the solution of the equation.