Chapter 7 Test: Exponents and Logarithms

Rewrite Exp and Logarithms

100

Rewrite as an exponential equation: log₉(1/81)=-2

9^-2 = 1/81

200

Rewrite as a logarithm: 3⁵ = 243

log₃243=5

300

Rewrite as an exponential equation: log8x = 2

10² = 8x

400

Rewrite as an exponential equation: ln(x - 5) = 6

e⁶ = x - 5

500

Rewrite as a logarithm: e^-5x =0.1

ln0.1 = -5x

Solve: Exponential

100

Solve: 5^(3x+1) = 5^{(4x - 5)}

x = 6

200

Solve: 6^x = 25

1.7965

300

Solve: 4^{(2x - 1)} = 64

x = 2

400

Solve: 9^{4x + 3} = (1/81)^{x - 10}

x = 17/6 or 2.833

500

Solve: 36^{(2x - 3)} = 216^{(2x - 4)}

x = 3

Solve: Logarithm

100

Solve: log₃₆x=(1/2)

x = 6

200

Solve: log₇(3x - 1) = log₇(6x + 20)

x = -7

300

Solve: log₄4x + log₄3 = log₄48

x = 4

400

Solve: log₃(x² + 5x) = log₃(3x + 24)

x = -6, x = 4

500

Solve: log₄x + log₄(x - 6) = 2

x = 8

Solve: e and ln

100

Solve: 4 + e^x = 19

2.7081

200

Solve: ln(4x) = 3

5.0214

300

Solve: e^2x + 1 = 55

1.9945

400

Solve: ln(x+2)=3

18.0855

500

Solve: ln(2x + 5)³ = 12

24.7991

Solve: Word Problems

100

Ms. Watson received a job as a teacher with a starting salary of $38,000. According to her contract, she will receive a 1.5% increase in her salary every year. How much will she be earning in 8 years? 

A = a(1 + r)^t

$42,806.72

200

A computer valued at $6500 depreciates at the rate of 14.3% per year. Find the value after three years. 

A = a(1- r)^t

$4,091.25

300

An investment of $17,500 increases at a annual rate of 7.2% compounded monthly. Find the value of the investment after 30 years.

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

$150,768.67

400

If $500 is deposited in an account paying 5.25% interest compounded continuously, find the amount in the account after 3 years.

A=Pe^rt

$585.29

500

If you deposit $800 into a savings account paying 4.5% interest compounded continuously, how long would you have to wait until you have at least $2000 in the account? 

A = Pe^rt

about 20.36 years