Question 1106522
<font color="black" face="times" size="3">Rule: a^b*a^c = a^(b+c)


Using that rule, we can say that
5^1*5^2 = 5^(1+2)
and by extension
5^1*5^2*5^3 = 5^(1+2)*5^3 = 5^(1+2+3)
and so on


For the entire left side, we can condense
5^1*5^2*5^3* ... * 5^k
into
5^(1+2+3+...+k)
using that rule mentioned above


Therefore we have 
5^1*5^2*5^3* ... * 5^k = 5^171
turn into
5^(1+2+3+...+k) = 5^171


The bases are both equal to 5, so the exponents must be equal
In other words,
1+2+3+...+k = 171


The left side can be found using the formula *[Tex \Large \displaystyle \sum_{m=1}^{k}m = 1+2+3+\ldots k = \frac{k(k+1)}{2}]


Which is why 
1+2+3+...+k = 171
becomes
k*(k+1)/2 = 171


Now solve for k
k*(k+1)/2 = 171
k*(k+1) = 171*2
k^2+k = 342
k^2+k-342 = 0
(k+19)(k-18) = 0
k+19 = 0 or k-18 = 0
k = -19 or k = 18


We only consider positive k values. So toss out k = -19
The only solution is k = 18
Note how 1+2+3+...+k = 1+2+3+...+18 = 171


If you want to see it more expanded out, then,
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18 = 171


Final Answer: k = 18
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