# How do you solve lnx+ln(x+4)=ln5?

Oct 31, 2016

Use the rule $\log \left(a\right) + \log \left(b\right) = \log \left(a \times b\right)$ to start the solving process.

$\ln \left(x \left(x + 4\right)\right) = \ln 5$

$\ln \left({x}^{2} + 4 x\right) = \ln 5$

We can now use the property $\log a = \log b \to a = b$ to solve.

${x}^{2} + 4 x = 5$

${x}^{2} + 4 x - 5 = 0$

$\left(x + 5\right) \left(x - 1\right) = 0$

$x = - 5 \mathmr{and} 1$

However, $x = - 5$ is extraneous, since the domain of $y = \ln x$ is $x > 0$.

The only solution is hence $x = 1$.

Hopefully this helps!

Oct 31, 2016

$\ln \left(x\right) + \ln \left(x + 4\right) = \ln \left(5\right)$

$\ln \left(x \left(x + 4\right)\right) = \ln \left(5\right)$

$x \left(x + 4\right) = 5$

${x}^{2} + 4 x = 5$

${x}^{2} + 4 x - 5 = 0$

Factorise:

$\left(x - 1\right) \left(x + 5\right) = 0$

This means that:

x is equal to 1 as it can't be -5 . This is your answer.