# How do you solve log_5(x+1) - log_4(x-2) = 1?

Dec 28, 2017

I wanted to provide a graphical approach to this problem...

The answer to this problem is where the function, $y = {\log}_{5} \left(x + 1\right) - {\log}_{4} \left(x - 2\right)$ interects with $y = 1$:

Hence from this graph we see that $x \approx 2.787$ is a solution for this...

Now i understand that this is not the most ellegant solution, and not the traditional method, so i want to hopefully start a more rigerouse approach:

We could possibly use a very forgotten identity in logarithms...

${\log}_{\alpha} \beta \equiv \frac{{\log}_{\phi} \beta}{{\log}_{\phi} \alpha}$

If you have never seen this before, its a good idea to add this to your inventory of identities...

Applying this...

$\implies \log \frac{x + 1}{\log} 5 - \log \frac{x - 2}{\log} 4 = 1$

$\implies \log 4 \cdot \log \left(x + 1\right) - \log 5 \cdot \log \left(x - 2\right) = \log 4 \cdot \log 5$

$\implies \log \left({\left(x + 1\right)}^{\log 4}\right) - \log \left({\left(x - 2\right)}^{\log 5}\right) = \log 4 \cdot \log 5$

$\implies \log \left({\left(x + 1\right)}^{\log 4} / {\left(x - 2\right)}^{\log 5}\right) = \log 4 \cdot \log 5$

$\implies {\left(x + 1\right)}^{\log 4} / {\left(x - 2\right)}^{\log 5} = {10}^{\log 4 \cdot \log 5}$

$\implies {\left(x + 1\right)}^{\log 4} = {10}^{\log 4 \cdot \log 5} \cdot {\left(x - 2\right)}^{\log 5}$

Now maybe we can use binomial epxansion and other approximations, but i hope this was a good start to one apporach, but maybe there are more simple approaches, hopefully someone else can provide such a method!