Question #5bb0e

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Answer 1 · 2017-09-20T21:13:24

k=2

in more detail:
#5^(k+1)=x#
#(5 * x) + x - 750 = 0#
#6 * x = 750 #
#->x=125 #
#5^(k+1)=125#
#->k=2#

Answer 2 · 2017-10-31T12:46:27

Taking to where you can take over

Just to make things a bit more strait forward:

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

#5^(m+3)+5^m=750#

#5^m(5^3+1)=750#

#5^m=750/126#

taking logs

#mln(5)=ln(750)-ln(126)#

#m= (ln(750)-ln(126))/ln(5) #

Thus #k-1= (ln(750)-ln(126))/ln(5)#

I will let you do the rest.