Solving exponential equations

Section 7.2 Solving exponential equations

Solve

 $3^x=9^4$  

• Notice the bases are not the same so you cannot set the exponents equal yet

• Look at the expressions and see if you can rewrite them so the bases are the same

• In the beginning it will typically mean rewriting the larger base in terms of the smaller base

•  $3^x=9^4$  

•  $3^x=\left(3^2\right)^4$  

• Simplify the right side by using the power to a power property

•  $3^x=3^8$  

• Now you can set the exponents equal since the bases are the same

• x = 8

Solve

 $2^{5x}=4^{2x-1^{ }}$  

•  $2^{5x}=\left(2^2\right)^{2x-1^{ }}$  

• Be careful with your power to a power property.  Here its 2(2x - 1) = 4x - 2

•  $2^{5x}=2^{4x-2^{ }}$  

• 5x = 4x - 2

• x = -2

Multiple Choice

Solve

 $4^x=64^3$  

• 3

• 9

• 12

• none of these

Answer with work

•  $4^x=64^3$  

•  $4^x=\left(4^3\right)^3$  

•  $4^x=4^9$  

• x = 9

Fill in the Blank

Solve

 $3^{2x}=9^{5x-4^{ }}$  

Answer with work

•  $3^{2x}=9^{5x-4^{ }}$  

•  $3^{2x}=\left(3^2\right)^{5x-4^{ }}$  

•  $3^{2x}=3^{10x-8^{ }}$  

• 2x = 10x - 8

• -8x = -8

• x = 1

Fill in the Blank

Solve

 $4^{2x+6^{ }}=64^{2x-4^{ }}$  

Answer with work

•  $4^{2x+6^{ }}=64^{2x-4^{ }}$  

•  $4^{2x+6^{ }}=\left(4^3\right)^{2x-4^{ }}$  

•  $4^{2x+6^{ }}=4^{6x-12^{ }}$  

• 2x + 6 = 6x - 12

• 18 = 4x

•  $x\ =\ \frac{9}{2}$  

Solve

 $64^{c-2^{ }}=32^{2c}$  

• Here you will need to rewrite both sides of the equation

•  $\left(2^6\right)^{c-2^{ }}=\left(2^5\right)^{2c}$  

•  $2^{6c-12^{ }}=2^{10c}$  

• 6c - 12 = 10c

• -12 = 4c

• c = -3

Solve

 $5^{3-2x^{ }}=\frac{1}{625}$  

• Remember your negative exponent properties?

•  $\frac{1}{625}=625^{-1}$  

•  $5^{3-2x^{ }}=625^{-1}$  

•  $5^{3-2x^{ }}=\left(5^4\right)^{-1}$  

•  $5^{3-2x^{ }}=5^{-4}$  

• 3 - 2x = -4

• -2x = -7

•  $x\ =\ \frac{7}{2}$  

Teen spending is expected to grow 3.5% annually from $79.7 billion in 2006.  

• A. Write the equation to model teen spending.

•  $A\left(t\right)=79.7\left(1+.035\right)^t$  in billion $

•  $A\left(t\right)=79.7\left(1.035\right)^t$  

• B.  Find the expected amount of money teens will spend in 2021.  Round to the nearest hundredth

•  $A\left(15\right)=79.7\left(1.035\right)^{15}$  

•  = $133.53 billion 

In 2010, the population of Phoenix was 1,321,045. By 2019, it was estimated at 1,512,986. Write an exponential function that could be used to model the population of Phoenix. Write t in terms of the numbers of years since 2010.

• Known: initial amount a = 1,321,045, final amount A(t) = 1,512,986, and time periods t = 9

• Unknown: growth rate 1 + r

• a = 1,321,045, A(t) = 1,512,986, and t = 9

•  $A\left(t\right)=a\left(1+r\right)^t$  

•  $1,512,986=1,321,045\left(1+r\right)^9$  

• Solve for 1 + r

•  $\frac{1,512,986}{1,321,045}=\left(1\ +\ r\right)^9$  

•  $\left(\frac{1,512,986}{1,321,045}\right)^{\frac{1}{9}}=\left(\left(1+r\right)^9\right)^{\frac{1}{9}}$  

• Rounded to the nearest tenth 1 + r = 1.02 which makes the model  $A\left(t\right)=1,321,045\left(1.02\right)^t$  

Multiple Choice

In 2010, the population of the town of Tisdale was 9,426. By 2017, it was estimated at 17,942. Write an exponential function that could be used to model the population of Tisdale. Write x in terms of the numbers of years since 2010.  

• $y=9,426\left(1.0963\right)^{x-7^{ }}$
  

• $y=1.0963\left(9,426\right)^x$

• $y=9,426\left(x\right)^{1.0963}$

• $y=9,426\left(1.0963\right)^x$

Answer with work

•  $17942=9426\left(1+r\right)^7$  

•  $\frac{17942}{9426}=\left(1+r\right)^7$  

•  $\left(\frac{17942}{9426}\right)^{\frac{1}{7}}=\left(\left(1+r\right)^7\right)^{\frac{1}{7}}$  

• 1 + r = 1.0963

•  $A\left(t\right)=9426\left(1.0963\right)^x$  

Fill in the Blank

Using the model from the last problem

 $A\left(t\right)=9426\left(1.0963\right)^x$  predict the population in 2022.  Remember x is years since 2010

An investment account pays 5.4% annual interest compounded quarterly. If $4000 is placed in this account, find the balance after 8 years.

• Use the compound interest formula with p = 4000, r = .054, n = 4, t = 8

•  $A\ =4000\left(1+\frac{.054}{4}\right)^{\left(4\cdot8\right)}$  

• $6143.56

Fill in the Blank

An investment account pays 4.6% annual interest compounded quarterly. If $6050 is placed in this account, find the balance after 6 years.

Homework

• In Aleks Section 7.2 due tomorrow 11:59 pm

• Wednesday we are meeting to work on Albert.io assignment SAT Prep: Equivalent expressions due Wednesday night 11:59 pm