Practice with Exponential Expressions and Equations Topic Index | Algebra2/Trig Index | Regents Exam Prep Center

Choose the best answer for each question.
(If not specified, round answer to the nearest thousandth when needed.)

 1. Solve for x: Choose: -7 3 7 Explanation Since the bases are the same, set the exponents equal and solve for x.

 2. Solve for x: Choose: -8 -6 6 Explanation Both 8 and 16 can be expressed as powers of 2. 8 = 2^3 16 = 2^4 Rewrite the problem using the base of 2 for each side. 2^(3x + 6) = 2^4x Set the exponents equal and solve for x.

 3. Solve for x: Choose: -3 -8   8 Explanation Express 81 as a power of 9 and rewrite the problem. 9^(3x) = 9^(2x-8) Set the exponents equal and solve for x.

 4. Solve for x: Choose: 0.699 0.698  0.545 Explanation A common base cannot be found for this problem. Solve using logarithms. log 10^x = log 5 x log 10 = log 5 x = log 10 / log 5 Estimate answer using calculator.

 5. Solve for x: Choose: 0.834 -0.834  -0.631 Explanation Use logarithms to solve. log (1/3)^x = log 2 x log (1/3) = log 2 x = log 2 / log (1/3) Estimate answer using calculator.

 6. Solve for x: Choose: -0.631 0.631 0.834 Explanation Use logarithms to solve. log 3^(-x+2) = log 18 (-x+2) log 3 = log 18 -x+2 = log 18 / log 2 Solve for x and estimate answer using calculator.

 7. Solve for x: Choose: 1.787 0.257 -2 Explanation A common base can be achieved in this problem. 1/9 = 1/3² = 3^(-2) Rewrite the problem. 3^x = 3^(-2) x = -2

 8. Simplify: Choose: 2 6 8 Explanation Move the 3 so that you can utilize the composition of the inverse functions. 3 is really the exponent of the two. e^(ln 2^3) = e^(ln 8) = 8

 9. Solve for x (to the nearest tenth): Choose: 0.3 0.4 0.7 Explanation Take ln of both sides. ln(e^(2x+3)) = ln 40 2x + 3 = ln 40 2x = ln (40) - 3 x = (ln (40) - 3) / 2 = .3444397271

 10. Solve for t (to the nearest hundredth): Choose: 1.43 4.19 4.20 Explanation Divide each side by 12 and solve using ln. e^.34t = 50/12 ln(e^.34t) = ln(50/12) .34t = ln(50/12) t=(ln(50/12)/.34)=4.197401046

 11. Simplify:         ln(e) Choose: 0 1 e Explanation ln(e) = ln(e^1) = ln(exp(1)) = 1. Remember that ln x and e^x are inverse functions. Under composition, inverse functions return the starting value.

 12. Solve for x: Choose: ln 5 -3 both ln 5 and -3 Explanation Factor: (e^x-5)(e^x+3)=0 Set each factor equal to 0. e^x - 5 = 0 tells us that e^x = 5 e^x + 3 = 0 tells us that e^x = -3 Since e^x is never equal to -3, we eliminate this possible solution. Now solve for x by taking ln of both sides: e^x = 5 ln(e^x) = ln 5 x = ln 5

 Topic Index | Algebra2/Trig Index | Regents Exam Prep Center Created by Donna Roberts