# Question #5bb0e

Sep 20, 2017

k=2

#### Explanation:

in more detail:
${5}^{k + 1} = x$
$\left(5 \cdot x\right) + x - 750 = 0$
$6 \cdot x = 750$
$\to x = 125$
${5}^{k + 1} = 125$
$\to k = 2$

Oct 31, 2017

Taking to where you can take over

#### Explanation:

Just to make things a bit more strait forward:

Set $m = k - 1$ then we have:

${5}^{m + 3} + {5}^{m} = 750$

${5}^{m} \left({5}^{3} + 1\right) = 750$

${5}^{m} = \frac{750}{126}$

taking logs

$m \ln \left(5\right) = \ln \left(750\right) - \ln \left(126\right)$

$m = \frac{\ln \left(750\right) - \ln \left(126\right)}{\ln} \left(5\right)$

Thus $k - 1 = \frac{\ln \left(750\right) - \ln \left(126\right)}{\ln} \left(5\right)$

I will let you do the rest.