LOGARITHM PROBLEMS WITH SOLUTIONS

Problem 1 :

Solve for x :

log2x - log25 = 2 + log23

Solution :

log2x - log25 = 2 + log23

Use the fundamental law of logarithm.

log2(x/5) = 2 + log23

Subtract log23 from each side.

log2(x/5) - log23 = 2

log2(x/15) = 2

Convert to exponential form.

x/15 = 22

x/15 = 4

Multiply each side by 15.

x = 60

Problem 2 :

Find integer values of m and n for which 

m - nlog32 = 10log96

Solution :

m - nlog32 = 10log96

Use the property of logarithm.

m - nlog32 = 10/log69

m - nlog32 = 10/log632

Use the fundamental law of logarithm.

m - nlog32 = 10/2log63

Simplify.

m - nlog32 = 5/log63

m - nlog32 = 5log36

m - nlog32 = 5log3(3 ⋅ 2)

Use the fundamental law of logarithm.

m - nlog32 = 5[log33 + log32]

m - nlog32 = 5[1 + log32]

Use the distributive property.

m - nlog32 = 5 + 5log32

Equate the constant terms and coefficients of like terms.

m = 5

-n = 5

n = -5

Problem 3 :

Solve the simultaneous equations :

log26x = 1 + 2log2y

1 + log6x = log6(15y - 25)

Solution :

log26x = 1 + 2log2y ----(1)

1 + log6x = log6(15y - 25) ----(2)

(1) :

log26x = 1 + 2log2y

log26x = log2+ 2log2y

log26x = log22 + log2y2

log26x = log2(2y2)

6x = 2y2 ----(3)

(2) :

1 + log6x = log6(15y - 25)

log66 + log6x = log6(15y - 25) 

log6(6x) = log6(15y - 25) 

6x = 15y - 25

From (3), substitute 2y2 for 6x in (4).

2y2 = 15y - 25

2y2 - 15y + 25 = 0

Factor and solve.

2y2 - 10y - 5y + 25 = 0

2y(y - 5) - 5(y - 5) = 0

(y - 5)(2y - 5) = 0

y = 5 or y = 5/2

Substitute 5 and 5/2 for y in (3).

6x = 2(5)2

6x = 2(25)

6x = 50

x = 50/6

x = 25/3

6x = 2(5/2)2

6x = 2(25/4)

6x = 25/2

x = 25/(2 ⋅ 6)

x = 25/12

Therefore, 

x = 25/3, y = 5

x = 25/12, y = 5/2

Problem 4 :

Solve for x :

log2(x + 3) + log2(x - 3) = 4

Solution :

log2(x + 3) + log2(x - 3) = 4

Use the fundamental law of logarithm.

log2[(x + 3)(x - 3)] = 4

Convert to exponential form.

(x + 3)(x - 3) = 24

x2 - 32 = 16

x2 - 9 = 16

x2 = 25

√x2 = √25

x = ± 5

x = -5 or 5

But, x = -5 will not work with the given equation.

Because, x = -5 makes the arguments negative in the logarithms. Since logarithm is defined only for positive value in argument, x = -5 is not a solution.

So, x = 5 is the solution to the above equation.

Problem 5 :

Prove that :

logr2x = (1/2)logrx

Solution :

logr2x ----(1)

(1/2)logrx ----(2)

(1) :

logr2x

Use the property of logarithm.

= 1/(logxr2)

1/(2logxr)

= 1(logrx)/2

= (1/2)logrx

= (2)

So, (1) = (2). 

logr2x = (1/2)logrx

Hence, proved.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Multiplicative Inverse Worksheet

    Jan 19, 22 09:34 AM

    Multiplicative Inverse Worksheet

    Read More

  2. Multiplicative Inverse

    Jan 19, 22 09:25 AM

    Multiplicative Inverse - Concept - Examples

    Read More

  3. Graphing Linear Functions Worksheet

    Jan 19, 22 08:24 AM

    Graphing Linear Functions Worksheet

    Read More