heres an example problem For the given situation, P = 2100, r = 0.006, and k = 4.Now, substitute 0.006 for r, 4 for k, and 2100 for P.A=P(1-r/k)^ktA=2100(1+0.006/4)^4trewrite the base with an exponent of tA=2100(1+0.006/4)^4tA=2100((1+0.006/4)^4)^tThis gives an exponential model of the form y = ab^x , where a = 2100 and b =(1+0.006/4)^4=1.0060, rounding to four decimal places. The corresponding linear model for the semi-log plot y = ( log b) x + log ca can be expressed using the natural logarithm as follows.Simplify further, rounding to four decimal places.y=(logcb) x+logcay=(ln1.0060)x+ln2100 substitute y=0.0060x+7.6497. simplify thus the equation corresponding linear model for the semi log plot is y=0.0060x+7.6497 so using this example do my problem

The formula for the amount A in an account after t years is as shown, where P is the principal, or amount invested into the account, and r is the annual interest rate (written as a decimal) that is compounded k times a year for t years. Suppose Celia invests $1700 at 2.6% annual interest compounded quarterly, that is, compounded 4 times a year. Determine the equation of the corresponding linear model for the semi-log plot. A = P 1 + K The equation is y = (Use integers or decimals for any numbers in the expression. Round to four decimal places as needed.)