1001 Algebra Problems.PDF

779. c.We rewrite both sides of the equation as a
     power of 2, then equate the exponents and
     solve for x:

4 x+1= (^12 )^2 x

(2^2 )x+1= (2–1)^2 x

2 2(x+1)= 2–2x

2(x+ 1) = –2x

2 x+ 2 = –2x

4 x= –2

x= –^12 

780. b.Factor the left side of the equation, set each
     factor equal to zero, and solve for x, as follows:

x 3 x+ 5 3 x= 0

3 x(x+ 5) = 0

3 x= 0 or x+ 5 = 0

Since 3x= 0 has no solutions, we conclude that

the only solution is x= –5.

781. b.We rewrite both sides of the equation as a
     power of 10, then equate the exponents and
     solve for x using the quadratic formula:

(10x+1)^2 x= 100

102 x(x+1)= 10^2

2 x(x+ 1) = 2

2 x^2 + 2x= 2

2 x^2 + 2x– 2 = 0

x^2 + x– 1 = 0

x= =

782. c.We rewrite both sides of the equation as a
     power of 2, then equate the exponents and
     solve for x using the quadratic formula:

2 x2 = 8

2 x+1= 2^3

x+ 1 = 3

x= 2

x= 4

783. c.Factor the left side of the equation, set each
     factor equal to zero, and solve for x:

2 x^2 ex– 7xex+ 6ex= 0

ex(2x^2 – 7x+ 6) = 0

ex(2x– 3)(x– 2) = 0

ex= 0 or 2x– 3 = 0 or x– 2 = 0

Since ex= 0 has no solutions, we conclude that

the solutions are ^32 and 2.

784. b.Factor the left side of the equation, set each
     factor equal to zero, and solve for x:

e^2 x+ 5ex– 6 = 0

(ex)^2 + 5(ex) – 6 = 0

(ex+ 6)(ex– 1) = 0

ex+ 6 = 0 or ex– 1 = 0

ex= – 6 = 0 or ex= 1

Since exis always positive, the first equation

has no solution. However, the second equation

is satisfied when x= 0.

Set 50 (Page 118)

785. 
     

     d.Finding xsuch that log 3 27 = xis equivalent
     to finding xsuch that 3x= 27. Since 27 = 3^3 ,
     the solution of this equation is 3.

     
     


786. 
     

     b.Finding xsuch that log (^3) ^19 = xis equivalent
     to finding xsuch that 3x= ^19 . Since ^19 = 3–2,we
     conclude that the solution of this equation is –2.

     
     


787. 
     

     c.Finding xsuch that log (^12) 8 = xis equivalent
     to finding xsuch that ^12 x= 8. Since 8 = 2^3
     and ^12 
     x
     = (2–1)x= 2–x, this equation is equiv-
     alent to 2^3 = 2–x, the solution of which is –3.

     
     


788. 
     

     c.Finding xsuch that log 7  7 = xis equivalent
     to finding xsuch that 7x=  7 . Since  7 =
     7 , we conclude that the solution of this
     equation is ^12 .

     
     

^12

–1 ±^5 

2

–1 ±1 – 4(1)( –1)

2(1)

ANSWERS & EXPLANATIONS–