Question #5bb0e

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Question #5bb0e

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Answer 1 · 2017-09-20T21:13:24

k=2

Explanation:

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

Answer 2 · 2017-10-31T12:46:27

Taking to where you can take over

Explanation:

Just to make things a bit more strait forward:

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

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

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

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

taking logs

$m ln (5) = ln (750) - ln (126)$ $mln(5)=ln(750)-ln(126)$

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

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

I will let you do the rest.

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Question #5bb0e

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