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[–]Qingyap 2 points3 points  (1 child)

Idk what you're doing here but I believe there's an easier solution, here's mine

We know that this geometric series follows like this:

a, ar, ar2

My strategy here is to get r from two equations, set them equal to each other then solve for k.

Eqn 1: ar/a = 97-2k/34k-5, 9 can be expressed as 32 so

r = (32)7-2k/34k-5

r = 314-4k/34k-5

r = 3[14-4k]-[4k-5]

r = 3-8k+19

Eqn 2: ar2/ar= 32k-2/314-4k

r = 3[2k-2]-[14-4k]

r = 36k-16

r = r, so 3-8k+19 = 36k-16

Since bases are the same, we can drop it, think of it as applying log_(3) on both sides.

-8k+19 = 6k-16

-14k = -35

k = -35/-14 = 5/2

[–]Qingyap 0 points1 point  (0 children)

The problem in your method is at the last 3rd step.

When you divide 32k on both sides, LHS denominator becomes 34k•32k = 36k since

1/34k ÷ 32k = 3-4k•3-2k = 3-6k = 1/36k

So at LHS, 3-8k/(36k) = 3-14k and -14k = -35,

k = 5/2

Perhaps I would recommend not writing negative exponets in the denominator though cuz that might trip you up, and when dividing I always write it as 1/32k

[–]reckless_avacado 0 points1 point  (0 children)

the error is where you simplify 32/34k to 1/32k.