W5A2 - Exponential Equations

What would be the first easiest action to take solving the equation  $2^{5x}\ =\ 2^{3\left(x+4\right)}$  

• Divide both sides by 2

• Subtract 4 from both sides

• Equate the exponents 5x and 3(x+4)

• Take log of both sides

Which is a possible first step one would take solving the equation  $3^{2x}\ =\ 27^{\left(x-4\right)}$  

• Equate the exponents 2x and (x-4)

• Divide both sides by 3

• Rewrite 27 as a power of 3

• Distribute (x-4)

In solving the exponential equation  $4^{2\left(x+3\right)}\ =4^{\left(1-x\right)}$  , why would I equate the exponents?

• Because they are powers and should be equated

• I can see that they are equal

• Because the bases are the same

• If I take log of both sides I will end up with that equation

What would you possibly do first solving the equation  $9^{2x}\ =\ 9^{20}$  

• Take log of both sides

• Divide both sides by 2

• Equate the exponents

• Divide both sides by 9

• Divide both sides by 20

Solve:

3^x-2 = 1

• 2

• 3

• -2

• 0

Solve: 7^-x = 49

• x = 1
• x = -1
• x = 2
• x = -2

Tags

CCSS.HSA.CED.A.1

CCSS.HSF.BF.B.5

Solve 3³ = 3^{4x + 2}

• x = ¼
• x = -¼
• x = ½
• x = -½

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.A.1

To solve 8 = 2^5x+7, you would need to re-write 8 as what base?

• 8
• 4
• 2
• Cannot be determined

Tags

CCSS.8.EE.A.1

Solve for a:
5^{a + 2} = 1/125

• -1
• -6/7
• -5
• 4

Tags

CCSS.HSF.LE.A.4

81^3-x = (1/3)^5x-6

• 6
• -6
• 9/2
• 3/2

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.A.1

CCSS.HSF.LE.A.4