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1. Five thousand dollars compounded annually at an x% interest rate takes six years to double. At the same interest rate, how many years will it take $300 to grow to$9600?

2. Find all values of C such that 3^2c+1=28*3^c-9 . If you find more than one value of, then list your values in
increasing order, separated by commas.

Mar 25, 2019

#1
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1. The first statement shows that 5000 + x/100 * 5000 + (x/100)^2 * 5000 ... (x/100)^6 *5000 = 10000. I believe you can solve that. Then, plug it on the 300 dollars, which will look like 300 + x/100 * 300 + (x/100)^2 * 300 .... (x/100)^n * 300 = 9600, where n is number you are looking for.

2. Unclear of what you are asking for. If you are asking for $$3^{2c+1} = 28\cdot 3^{c-9}$$, the answer is in some really ugly ln form. If you are asking for $$3^2 c + 1 = 28 \cdot 3^c -9,$$ then it is also really ugly. Please specify what your equation is asking for.

Mar 25, 2019
#2
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Thans for helping. I think, I have leads for the second one.

The solution shows this:

3^(2c+1)=28*(3^c)-9

3^(2c)*3^1 = 28*3^c - 9

3(3^c)^2 - 28(3^c) + 9 = 0

let 3^c = x

3x^2 - 28x + 9 = 0

(3x - 1)(x - 9) = 0

x = 1/3 or x = 9

if x = 1/3

3^c= 1/3 = 3^-1

c = -1

if x = 9

3^c = 9 = 3^2

c = 2

Mar 25, 2019