SOLVING EXPONENTIAL EQUATIONS USING LOGARITHMS WORKSHEET

Problem 1 :

Solve for x :

4x - 5 = 16

Problem 2 :

Solve for x :

32x - 4 = 93x + 6

Problem 3 :

Solve for m :

163m - 6 = 1

Problem 4 :

Solve for x. Round your answer to the nearest ten-thousandth.

3x = 17

Problem 5 :

Solve for y. Round your answer to the nearest ten-thousandth.

20y = 56

Problem 6 :

Solve for k. Round your answer to the nearest ten-thousandth.

5 ⋅ 186k = 26

Problem 7 :

Solve for x. Round your answer to the nearest ten-thousandth.

ex - 1 - 5 = 5

Problem 8 :

Solve for p. Round your answer to the nearest ten-thousandth.

-e6 - 9p + 5 = -48.4

Problem 9 :

Solve for r.

8r + 4  42r - 10 = 16r + 1

Problem 10 :

Ax + B = CDx

Given that A, B, C and D are all real numbers greater than zero. Solve for x in terms of A, B, C and D.

tutoring.png

Answers

1. Answer :

4x - 5 = 16

The base on the left side is 4 and that of on the right side is 16.

2 = 22

16 = 24

Since 2 and 16 are the powers of 2, we can take logarithm with base 2 on both sides of the given equations and solve for x.

log2(4x - 5) = log2(16)

log2[(22)x - 5] = log2(24)

log2(22(x - 5)) = 4log2(2)

2(x - 5)log2(2) = 4log2(2)

2(x - 5)(1) = 4(1)

2x - 10 = 4

2x = 14

x = 7

2. Answer :

32x - 4 = 93x + 6

The base on the left side is 3 and that of on the right side is 9.

3 = 31

9 = 32

Since 3 and 9 are the powers of 3, we can take logarithm with base 3 on both sides of the given equations and solve for x.

log3(32x - 4) = log3(93x + 6)

(2x - 4)log3(3) = (3x + 6)log3(9)

(2x - 4)(1) = (3x + 6)log3(32)

(2x - 4)(1) = (3x + 6) ⋅ 2log3(3)

2x - 4 = (3x + 6) ⋅ 2(1)

2x - 4 = 6x + 12

-4x = 16

x = -4

3. Answer :

163m - 6 = 1

The base on the left side is 16 and that of on the right side is 1.

16 = 42

1 = 40

Since 16 and 1 are the powers of 4, we can take logarithm with base 4 on both sides of the given equations and solve for m.

log4(163m - 6) = log1(1)

(3m - 6)log4(16) = 0

(3m - 6)log4(42) = 0

(3m - 6) ⋅ 2log4(4) = 0

(3m - 6) ⋅ 2(1) = 0

6m - 12 = 0

6m = 12

m = 2

4. Answer :

3x = 17

The base on the left side is 3 and that of on the right side is 17.

Since 3 and 17 are the not the powers of a common number, we can take common logarithm (with base 10) on both sides of the given equations and solve for x.

ln(3x) = ln(17)

⋅ ln(3) = ln(17)

Using calculator,

 2.5789

5. Answer :

20y = 56

The base on the left side is 20 and that of on the right side is 56.

Since 20 and 56 are the not the powers of a common number, we can take natural logarithm on both sides of the given equations and solve for y.

ln(20y) = ln(56)

⋅ ln(20) = ln(56)

Using calculator,

 1.3437

6. Answer :

5 ⋅ 186k = 26

Divide both sides by 5.

186k = 5.2

The base on the left side is 18 and that of on the right side is 5.2.

Since 18 and 5.2 are the not the powers of a common number, we can take natural logarithm both sides of the given equations and solve for k.

ln(186k) = ln(5.2)

(6k)ln(18) = ln(5.2)

 1.3437

7. Answer :

ex - 1 - 5 = 5

Add 5 to both sides.

ex - 1 = 10

The base on the left side is e and that of on the right side is 10.

Since e and 10 are the not the powers of a common number, we can take natural logarithm on both sides of the given equations and solve for x.

ln(ex - 1) = ln(10)

(x - 1)ln(e) = ln(10)

(x - 1)(1) = ln(10)

x - 1 = ln(10)

x = ln(10) + 1

 3.3026

8. Answer :

-e6 - 9p + 5 = -48.4

Subtract 5 from both sides.

-e6 - 9p = -53.4

Multiply both sides by -1.

e6 - 9p = 53.4

The base on the left side is e and that of on the right side is 53.4.

Since e and 53.4 are the not the powers of a common number, we can take natural logarithm on both sides of the given equations and solve for x.

ln(e6 - 9p) = ln(53.4)

(6 - 9p)ln(e) = ln(53.4)

(6 - 9p)(1) = ln(53.4)

6 - 9p = ln(53.4)

-9p = ln(53.4) - 6

 0.2247

9. Answer :

8r + 4  42r - 10 = 16r + 1

8 = 23

4 = 22

16 = 24

Since the bases 8, 4 and 16 are the powers of 2, we can take logarithm with base 2 on both sides of the given equations and solve for r.

log2(8r + 4  42r - 10) = log2(16r + 1)

log2(8r + 4) + log2(42r - 10) = log2(16r + 1)

(r + 4)log2(8) + (2r - 10)log2(4) = (r + 1)log2(16)

(r + 4)log2(23) + (2r - 10)log2(22) = (r + 1)log2(24)

(r + 4) ⋅ 3log2(2) + (2r - 10) ⋅ 2log2(2) = (r + 1) ⋅ 4log2(2)

(r + 4) ⋅ 3(1) + (2r - 10) ⋅ 2(1) = (r + 1) ⋅ 4(1)

3(r + 4) + 2(2r - 10) = 4(r + 1)

3r + 12 + 4r - 20 = 4r + 4

7r - 8 = 4r + 4

3r = 12

r = 4

10. Answer :

Ax + B = CDx

Since the bases A and C are alphabets, we can take natural logarithm on both sides of the given equations and solve for x in terms A, B, C and D.

ln(Ax + B) = ln(CDx)

(x + B)ln(A) = (Dx)ln(C)

xln(A) + Bln(A) = Dxln(C)

xln(A) - Dxln(C) = -Bln(A)

x[ln(A) - Dln(C)] = -Bln(A)

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