Solving logarithms

You want to solve for x. Which of these is the correct "step 1"?

• $\log_78=\log_7x$

• $\log_742=\log_7x$

• $\log_748=\log_7\left(\frac{x}{6}\right)$

• $\log_754=\log_7x$

Finish solving for x

• x = 6

• x = 7

• x = 8

• x = 56

Tags

CCSS.HSA.REI.A.1

CCSS.HSA.REI.B.3

Continue solving for x. Which is the correct "step 2"?

• $64=x^3$

• $64^3=x$

• $x^{64}=3$

• $3^{64}=x$

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.A.1

CCSS.HSF.BF.B.5

CCSS.HSF.LE.A.4

Finish solving for x.

Fill in the blank in the box below

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.A.1

CCSS.HSA.REI.B.3

Solve for x.
 $\log_52+\log_5x=\log_520$  
x = _______

Tags

CCSS.HSA.REI.A.1

CCSS.HSF.BF.B.5

Solve for x.
 $2\log_410=\log_4\left(8x\right)+\log_4\left(5\right)$  

• x = 2.5

• x = 0.5

• x = 1.538

• x = 11.875

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.A.1

CCSS.HSF.BF.B.5

CCSS.HSF.LE.A.4

 $\log_4x^2=\log_43+\log_4\left(x+6\right)$  

Which quadratic equation would you wind up solving when solving the log equation?

•  $x^2-3x-18=0$  

•  $x^2+3x+6=0$  

•  $x^2-3x-6=0$  

•  $x^2-x-9=0$  

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.B.4

log₄(3x-1)=log₄(2x+3)

• 4

• 3

• 1

• 8

Tags

CCSS.HSA.CED.A.1

CCSS.HSF.BF.B.5

log₈(4x+4)=2

• 15

• 12

• 10

• 3

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.A.1

CCSS.HSF.BF.B.5

Solve for x:
4^2x + 1 = 17

• x = 1
• x = -1
• x = 2
• x = -2

Tags

CCSS.7.EE.B.4A