WORKSHEET ON LOGARITHMIC EQUATIONS

Problem 1 :

Solve for x :

log6(x + 2) = 4

Problem 2 :

Solve for y :

log4(y - 5) = 3

Problem 3 :

Solve for a :

Problem 4 :

Solve for z :

log4(z) + log4(z - 15) = 2

Problem 5 :

Solve for a in terms of b :

log10(a) + log10(b)log10(a + b)

Problem 6 :

Solve for m :

log14(m - 3) + log14(m + 2) = 1

Problem 7 :

Solve for y :

Problem 8 :

What are the restrictions on b, so that x has two real solutions in the following equation?

tutoring.png

Answers

1. Answer :

log6(x + 2) = 4

Convert the above equation to exponential form.

x + 2 = 46

x + 2 = 4096

Subtract 2 from both sides.

x = 4094

2. Answer :

log4(y - 5) = 3

Convert the above equation to exponential form.

y - 5 = 43

y - 5 = 64

Add 5 from both sides.

x = 69

3. Answer :

Convert the above equation to exponential form.

a = 23

a = 8

4. Answer :

log4[z(z - 15)] = 2

log4(z2 - 15z) = 2

Convert the above equation to exponential form.

z2 - 15z = 42

z2 - 15z = 16

Subtract 16 from both sides.

z2 - 15z - 16 = 0

Solve by factoring.

z2 - 16z + z - 16 = 0

z(z - 16) + 1(z - 16) = 0

(z - 16)(z + 1) = 0

z - 16 = 0  or  z + 1 = 0

z = 16  or  z = -1

The aruguments of logarithms are always positive or greater than zero.

z > 0

z - 15 > 0

z > 15

The common solution of the inequalities z > 0 and z > 15 is

z > 15

Since z > 15, the solution z = -1 can not be acccepted.

Therefore,

z = 16

5. Answer :

log10(ab) = log10(a + b)

If two logarithms are equal with the same base, then their arguments must be equal.

ab = a + b

Subtract b from both sides.

ab - a = b

a(b - 1) = b

6. Answer :

log14[(m - 3)(m + 2)] = 1

log14(m2 - m - 6) = log14(14)

In the equation above, sine the two logarithms are equal with the same base, their arguments can be equated.

m2 - m - 6 = 14

Subtract 14 from both sides.

m2 - m - 20 = 0

Solve by factoring.

m2 - 5m + 4m - 20 = 0

m(m - 5) + 4(m - 5) = 0

(m - 5)(m + 4) = 0

m - 5 = 0  or  m + 4 = 0

m = 5  or  m = -4

The aruguments of logarithms are always positive or greater than zero.

m - 3 > 0

m > 3

m + 2 > 0

m = -2

The common solution of the inequalities m > 3 and m > -2 is

m > 3

Since m > 3, the solution m = -4 can not be acccepted.

Therefore,

m = 5

7. Answer :

8. Answer :

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