1) If \(11^{5y}=\left(\frac{1}{121}\right)^{2y+3}\), find \(y\). Express your answer in simplest fractional form.

2) It's January 1st, and Jillian is starting a new savings plan. Her savings account charges a \( \$10\) monthly fee for any month where she does not have an ending balance of\( \$1,\!000\) or higher. At the beginning of each month, she deposits \(\$100\) into her account. At the end of the year, her bank awards her an interest at the rate of \(1\%.\) rounded to the nearest dollar, calculated on the ending balance.

After three years, what is her ending balance after interest?

Thx in advance

-wolf

Guest Aug 29, 2019

#1**+1 **

1) 11^(5y) = (1 /121)^(2y+3) {Note : 1/121 = 1/11^2 = 11^(-2) }

So we have

11^(5y) = (11^(-2) )^(2y + 3)

11^(5y) = (11)^(-4y - 6) we have the bases the same......solve for the exponents

5y = -4y - 6 add 4y to both sides

9y = -6 divide both sides by 9

y = -6/9 = -2/3

CPhill Aug 29, 2019

#2**+1 **

At the end of October of the first year she will have (100 - 10)*10 = 90 * 10 = $900 in her account

In November....she contributes $100 and she will have $1000 in her acct at the end of November

In December, she contributes $100 and she will have $1000 + $100 = $1100 in her acct at the end of the year

She will then make 1% interest on this so she will have an ending balance of 1100 (1.01) = $1111 in her account

The second year she adds $1200 to this = $2311 and she makes 1% interest on this = 2311 (1.01) =

$2334 at the end of the second year

And in the third year...she adds $1200 more to this = 3534 and she makes 1% interest on this and ends up with

3534(1.01) = $3569

CORRECTED ANSWER.......!!!!

CPhill Aug 29, 2019