Noah currently has an account balance of $2,976.45. he opened the account 11 years ago with a deposit of $1,963.45. if the interest compounds quarterly, what is the interest rate on the account?

The interest rate is 3.79% or 3.8%

Step-by-step explanation:

The formula to find how much money A we have at the end of t years  if we start with a principal or  amount of money P today and the interest is compounded n times in the years is:

A=P(1+ r/n)^(nt)   (1)

Here A = $2,976.45 the final amount

t = 11 years

P =  $1,963.45.

and n = 4 because the interest is compounded quarterly or 4 times in the year.

We the need to simplify  the interest rate r in (1) so:

A=P(1+ r/n)^(nt)   (1)

A/P = (1+ r/n)^(nt)

(A/P)^(1/nt) = (1+ r/n)

(A/P)^(1/nt) - 1  =  r/n

((A/P)^(1/nt) - 1  ) * n =  r

r = ((A/P)^(1/nt) - 1  ) * n

We replace values and solve doing parenthesis exponents first

r = ((2,976.45/1,963.45.)^(1/(4*11)) - 1  ) * 4

r= ((1.51^(1/44)) -1)*4

((1.51^(1/44)) -1)*4

r= (1.009499443  - 1)*4

r= 0.009499443  *4 = 0.037997773

That expressed in percentage multiplying by 100 is 3.79% or 3.8% interest

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

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step-by-step explanation: