Solving Equations Using Logarithms

Edexcel IAL WMA12/01/P2/June 2021/Q7 (Logarithm,​​ The Factor Theorem)

• Given that​​ 

3 log3 2x – 1= 2 +log3⁡14x – 25

show that​​ 

2x3 – 3x2 – 30x + 56 = 0 

(4)

• Show that –4 is a root of this cubic equation.​​ 

(2)​​ 

• Hence, using algebra and showing​​ each step of your working, solve​​ 

3log3⁡(2x – 1)  = 2 +log3⁡(14x – 25) 

(4)

SOLUTION

a- Usng the laws of logarithm.

3log3⁡2x – 1-log3⁡14x – 25 =2

log3⁡2x-13-log3⁡14x-25=2

Applying the division rule from the flash card.​​ 

log3⁡2x-1314x-25=2

Converting the log equation to expnential one (index form). Therefore, the base of log becomes the base of indices.

2x-1314x-25=32

2x-13=914x-25

2x-12x-12=126x-225

2x-14x2-4x+1=126x-225

8x3-8x2+2x-4x2+4x-1=126x-225

8x3-12x2+6x-1-126x+225= 0

8x3-12x2-120x+224=0

Taking 4 as a common factor.​​ 

2x3-3x2-30x+56=0

b-​​ 

fx=2x3-3x2-30x+56

Using the factor theorem.​​ 

If x= -4 is a root then f-4=0

f-4=2-43-3-42-30 -4+56

f-4=2-64-316+120+56

f-4=-128-48+176

f-4=-176+176=0

 f-4=0

Since​​ f-4​​ comes out to be 0 therefore, -4 is a root of the cubic equation.​​ 

c- ​​ Since​​ x=-4 is a root ,  x+4 is a factor of​​ fx. Using the long division method to find the other two roots of the​​ equation.​​ 

2x2

x+42x3-3x2-30x+58-2x3+8x2                            -11x2-30x                                  --11x2-44x                                        14x+56                                      14x +56-         -               -

2x3-3x2-30x+56=0

Hence,​​ 

x+42x2-11x+14=0

x+42x-7x-2=0

x=-4,    2x=7          x=2

x=72

Now, discarding those values of x that gives argument either negative or zero.​​ 

3log3⁡2x-1=2+log3⁡14x-28

x= -4→undefined

x=72,     x=2