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# 0410 ch 4 day 10

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• 1. 4.4 Exponential &amp; Logarithmic Equations Day 3Romans 15:13&#xA0;&#xA0;May the God of hope &#xFB01;ll you withall joy and peace in believing, so that by the powerof the Holy Spirit you may abound in hope.
• 2. \$2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly.
• 3. \$2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly. nt &#x239B;&#xF8EB; r &#x239E;&#xF8F6; A = P &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; n &#x23A0;&#xF8F8;
• 4. \$2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly. nt &#x239B;&#xF8EB; r &#x239E;&#xF8F6; A = P &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; n &#x23A0;&#xF8F8; 12t &#x239B;&#xF8EB; .035 &#x239E;&#xF8F6;4000 = 2000 &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; 12 &#x23A0;&#xF8F8;
• 5. \$2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly. nt &#x239B;&#xF8EB; r &#x239E;&#xF8F6; A = P &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; n &#x23A0;&#xF8F8; 12t &#x239B;&#xF8EB; .035 &#x239E;&#xF8F6;4000 = 2000 &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; 12 &#x23A0;&#xF8F8; &#x239B;&#xF8EB; .035 &#x239E;&#xF8F6;log 2 = 12t log &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; 12 &#x23A0;&#xF8F8;
• 6. \$2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly. nt &#x239B;&#xF8EB; r &#x239E;&#xF8F6; A = P &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; n &#x23A0;&#xF8F8; log 2 =t &#x239B;&#xF8EB; .035 &#x239E;&#xF8F6; 12t 12 log &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239B;&#xF8EB; .035 &#x239E;&#xF8F6; &#x239D;&#xF8ED; 12 &#x23A0;&#xF8F8;4000 = 2000 &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; 12 &#x23A0;&#xF8F8; &#x239B;&#xF8EB; .035 &#x239E;&#xF8F6;log 2 = 12t log &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; 12 &#x23A0;&#xF8F8;
• 7. \$2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly. nt &#x239B;&#xF8EB; r &#x239E;&#xF8F6; A = P &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; n &#x23A0;&#xF8F8; log 2 =t &#x239B;&#xF8EB; .035 &#x239E;&#xF8F6; 12t 12 log &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239B;&#xF8EB; .035 &#x239E;&#xF8F6; &#x239D;&#xF8ED; 12 &#x23A0;&#xF8F8;4000 = 2000 &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; 12 &#x23A0;&#xF8F8; t &#x2248; 19.8 years &#x239B;&#xF8EB; .035 &#x239E;&#xF8F6;log 2 = 12t log &#x239C;&#xF8EC; 1+ &#x239F;&#xF8F7; &#x239D;&#xF8ED; 12 &#x23A0;&#xF8F8;
• 8. A sum of money invested at a &#xFB01;xed rate with continuouscompounding tripled in 19 years. Find the rate.
• 9. A sum of money invested at a &#xFB01;xed rate with continuouscompounding tripled in 19 years. Find the rate. rt A = Pe
• 10. A sum of money invested at a &#xFB01;xed rate with continuouscompounding tripled in 19 years. Find the rate. rt A = Pe 19r 3 = 1e
• 11. A sum of money invested at a &#xFB01;xed rate with continuouscompounding tripled in 19 years. Find the rate. rt A = Pe 19r 3 = 1e ln 3 = 19r ln e
• 12. A sum of money invested at a &#xFB01;xed rate with continuouscompounding tripled in 19 years. Find the rate. rt A = Pe 19r 3 = 1e ln 3 = 19r ln e ln 3 =r 19
• 13. A sum of money invested at a &#xFB01;xed rate with continuouscompounding tripled in 19 years. Find the rate. rt A = Pe 19r 3 = 1e ln 3 = 19r ln e ln 3 =r 19 r &#x2248; 5.8%
• 14. If I o and I denote the intensity of light before andafter going through a material, and x is the distance infeet the light travels in the material, then &#x2212;1 &#x239B;&#xF8EB; I &#x239E;&#xF8F6; ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = x k &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8;where k is a constant depending on the type of material.
• 15. If I o and I denote the intensity of light before andafter going through a material, and x is the distance infeet the light travels in the material, then &#x2212;1 &#x239B;&#xF8EB; I &#x239E;&#xF8F6; ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = x k &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8;where k is a constant depending on the type of material. This is the Beer-Lambert Law
• 16. If I o and I denote the intensity of light before andafter going through a material, and x is the distance infeet the light travels in the material, then &#x2212;1 &#x239B;&#xF8EB; I &#x239E;&#xF8F6; ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = x k &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8;where k is a constant depending on the type of material. This is the Beer-Lambert Law Solve the equation for I
• 17. &#x2212;1 &#x239B;&#xF8EB; I &#x239E;&#xF8F6; ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = xk &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8;
• 18. &#x2212;1 &#x239B;&#xF8EB; I &#x239E;&#xF8F6; ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = xk &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8; &#x239B;&#xF8EB; I &#x239E;&#xF8F6;ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = &#x2212;kx &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8;
• 19. &#x2212;1 &#x239B;&#xF8EB; I &#x239E;&#xF8F6; ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = xk &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8; &#x239B;&#xF8EB; I &#x239E;&#xF8F6;ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = &#x2212;kx &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8; &#x2212; kx I e = Io
• 20. &#x2212;1 &#x239B;&#xF8EB; I &#x239E;&#xF8F6; ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = xk &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8; &#x239B;&#xF8EB; I &#x239E;&#xF8F6;ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = &#x2212;kx &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8; &#x2212; kx I e = Io &#x2212; kx I oe =I
• 21. &#x2212;1 &#x239B;&#xF8EB; I &#x239E;&#xF8F6; ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = xk &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8; &#x239B;&#xF8EB; I &#x239E;&#xF8F6;ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = &#x2212;kx &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8; &#x2212; kx I e = Io &#x2212; kx I oe =I or &#x2212; kx I = I oe
• 22. &#x2212;1 &#x239B;&#xF8EB; I &#x239E;&#xF8F6; ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = x This is just a variation ofk &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8; rt A = Ao e &#x239B;&#xF8EB; I &#x239E;&#xF8F6;ln &#x239C;&#xF8EC; &#x239F;&#xF8F7; = &#x2212;kx or &#x239D;&#xF8ED; I o &#x23A0;&#xF8F8; rt &#x2212; kx I y = Pe e = Io &#x2212; kx I oe =I (This is part of Example 10 or on page 364) &#x2212; kx I = I oe
• 23. HW #9It is literally true that you can succeed best andquickest by helping others to succeed. Napoleon Hill