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Which Equation? $1000 is invested in an account that accrues 12% annual interest. A radioactive isotope has a half-life of 12 hours. $500 is deposited.

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Presentation on theme: "Which Equation? $1000 is invested in an account that accrues 12% annual interest. A radioactive isotope has a half-life of 12 hours. $500 is deposited."— Presentation transcript:

1 Which Equation? $1000 is invested in an account that accrues 12% annual interest. A radioactive isotope has a half-life of 12 hours. $500 is deposited into an account with 10% annual interest compounded weekly. The population grows at a rate of 1.5% compounded continuously. The car was purchased for $25,000. Ten years later, it was worth $12,000. Find the depreciation rate. How long will it take to triple your money if you invest in an account with a 3% annual interest rate?

2 Section 5.7 Exponential Equations Changing Bases
Objective: To solve exponential equations and to change the base of logarithms.

3 The Change of Base Formula
𝑙𝑜𝑔 𝑏 𝑐= 𝑙𝑜𝑔 𝑎 𝑐 𝑙𝑜𝑔 𝑎 𝑏 The change of base formula can be used to find the value of the exponent if it is unknown. Determine the value of x: x=30 Rewrite using the definition of log: 𝑙𝑜𝑔 2 30=𝑥 Use the change of base formula to change to base 10: ≈4.9

4 The Change of Base Formula
𝑙𝑜𝑔 𝑏 𝑐= 𝑙𝑜𝑔 𝑎 𝑐 𝑙𝑜𝑔 𝑎 𝑏 Solve: 5 𝑥 = 𝑥 =17 𝑙𝑜𝑔 4 17=𝑥 𝑙𝑜𝑔 5 8=𝑥 𝑥= 𝑙𝑜𝑔 5 8= 𝑙𝑜𝑔8 𝑙𝑜𝑔5 𝑥= 𝑙𝑜𝑔 4 17= 𝑙𝑜𝑔17 𝑙𝑜𝑔4 𝑥≈1.29 𝑥≈2.04

5 Example 1 In 2016, there are about 7.4 billion people in the world. If the population grows at 1.95% per year, estimate the year when the population will be 10 billion people.

6 In 2016, there are about 7. 4 billion people in the world
In 2016, there are about 7.4 billion people in the world. If the population grows at 1.95% per year, estimate the year when the population will be 10 billion people. 10=7.4 ( ) 𝑛 *We need to find the exponent! 10=7.4 (1.0195) 𝑛

7 Log Both Sides: Change to Log Form: 10=7.4 (1.0195) 𝑛 10=7.4 (1.0195) 𝑛 1.35= (1.0195) 𝑛 1.35= (1.0195) 𝑛 𝑙𝑜𝑔1.35= 𝑙𝑜𝑔(1.0195) 𝑛 𝑙𝑜𝑔 =𝑛 𝑙𝑜𝑔1.35=𝑛 𝑙𝑜𝑔(1.0195) 𝑛= 𝑙𝑜𝑔1.35 𝑙𝑜𝑔(1.0195) 𝑛= 𝑙𝑜𝑔1.35 𝑙𝑜𝑔(1.0195) ≈15.5 ≈15.5

8 Example 2 Suppose you invest P dollars at an annual rate of 6% compounded continuously. How long does it take: To increase your investment by 50%? To double your money?

9 Suppose you invest P dollars at an annual rate of 6% compounded continuously. How long does it take:
To increase your investment by 50%? 1.5𝑃=𝑃 𝑒 .06𝑡 *Divide both sides by P 1.5= 𝑒 .06𝑡 *Rewrite in log form 𝑙𝑛1.5=.06𝑡 *Solve for t 𝑡= 𝑙𝑛 ≈6.76

10 Suppose you invest P dollars at an annual rate of 6% compounded continuously. How long does it take:
To double your money? 2𝑃=𝑃 𝑒 .06𝑡 *Divide both sides by P 2= 𝑒 .06𝑡 *Rewrite in log form 𝑙𝑛2=.06𝑡 *Solve for t 𝑡= 𝑙𝑛2 0.06 ≈11.55

11 Additional Example An investment of $500 is made at 3.6% annual interest. How long does it take to triple the investment: A) if it is compounded monthly? B) if it is compounded continuously?

12 𝒏=𝟑𝟎.𝟔 𝒚𝒆𝒂𝒓𝒔 3= (1.003) 12𝑛 𝑙𝑜𝑔 1.003 3=12𝑛 12𝑛= 𝑙𝑜𝑔3 𝑙𝑜𝑔(1.003)
An investment of $500 is made at 3.6% annual interest. How long does it take to triple the investment: A) if it is compounded monthly? 1500=500 ( ) 12𝑛 3= (1.003) 12𝑛 𝑙𝑜𝑔 =12𝑛 12𝑛= 𝑙𝑜𝑔3 𝑙𝑜𝑔(1.003) 𝒏=𝟑𝟎.𝟔 𝒚𝒆𝒂𝒓𝒔

13 1500=500 𝑒 .036𝑡 3= 𝑒 .036𝑡 𝑙𝑛3=.036𝑡 𝑡= 𝑙𝑛3 .036 =𝟑𝟎.𝟓 𝐲𝐞𝐚𝐫𝐬
An investment of $500 is made at 3.6% annual interest. How long does it take to triple the investment: B) if it is compounded continuously? 1500=500 𝑒 .036𝑡 3= 𝑒 .036𝑡 𝑙𝑛3=.036𝑡 𝑡= 𝑙𝑛 =𝟑𝟎.𝟓 𝐲𝐞𝐚𝐫𝐬

14 American Mathematics Competition
Sign up by Friday, November 18 in the library. AMC 10 – 10th grade and below, Tuesday, February 7 AMC 12 – 12th grade and below, Wednesday, Februrary 15 A question from the 2015 AMC 12 Test:

15 Homework: Page #1-31 odds, skip 23


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