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

Feb 16, 2017

I would go for $x = 2.8$ but this is only an approximation of the real value and also I suspect the base of the second log should be $5$ as well....good exercise though!

#### Explanation:

We can try changing base to the second log and write it as:
$\frac{{\log}_{5} \left(x - 2\right)}{{\log}_{5} \left(4\right)}$
so the equation becomes:
${\log}_{5} \left(x + 1\right) - \frac{{\log}_{5} \left(x - 2\right)}{{\log}_{5} \left(4\right)} = 1$
to simplify our evaluation let us write ${\log}_{5} \left(4\right) = k$:
${\log}_{5} \left(x + 1\right) - \frac{{\log}_{5} \left(x - 2\right)}{k} = 1$
rearranging:
$k {\log}_{5} \left(x + 1\right) - \left({\log}_{5} \left(x - 2\right)\right) = k$
using a first property of logs:
${\log}_{5} {\left(x + 1\right)}^{k} - \left({\log}_{5} \left(x - 2\right)\right) = k$
a second ne:
${\log}_{5} \left[{\left(x + 1\right)}^{k} / \left(x - 2\right)\right] = k$
and the defintion of log:
${\left(x + 1\right)}^{k} / \left(x - 2\right) = {5}^{k}$
remembering that ${\log}_{5} \left(4\right) = k$ we have:
${\left(x + 1\right)}^{k} / \left(x - 2\right) = 4$
and:
${\left(x + 1\right)}^{k} = 4 x - 8$
but...the problem is $k$....I got that:
${\log}_{5} \left(4\right) = k = 0.86135$
to keep on going I will try to approximate and write that $k = 1$...ok it is not right but let us try it.
we get:
${\left(x + 1\right)}^{1} = 4 x - 8$
rearranging:
$x + 1 = 4 x - 8$
$3 x = 9$
$x = \frac{9}{3} = 3 + E$
where $E$ is the error introduced by my approximation.

Let us test our result into the original equation:
${\log}_{5} \left(4\right) - {\log}_{4} \left(1\right) = 1$
$0.86135 - 0 = 1$ more or less....

I tried various values for the Error $E$ and I found that if $E = - 0.2$
we get that:
$x = 3 - 0.2 = 2.8$
giving in the test of our equation with $x = 2.8$:
${\log}_{5} \left(2.8 + 1\right) - {\log}_{4} \left(2.8 - 2\right) = 1$
$0.99 = 1$ I think it is ok....