SOLVING EXPONENTIAL EQUATIONS USING LOGARITHMS WORKSHEET

Problem 1 :

Solve for x :

4^{x - 5} = 16

Problem 2 :

Solve for x :

3^{2x - 4} = 9^{3x + 6}

Problem 3 :

Solve for m :

16^{3m - 6} = 1

Problem 4 :

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

3^x = 17

Problem 5 :

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

20^y = 56

Problem 6 :

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

5 ⋅ 18^6k = 26

Problem 7 :

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

e^{x - 1} - 5 = 5

Problem 8 :

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

-e^{6 - 9p} + 5 = -48.4

Problem 9 :

Solve for r.

8^{r + 4} ⋅ 4^{2r - 10} = 16^{r + 1}

Problem 10 :

A^{x + B} = C^Dx

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.

Answers

1. Answer :

4^{x - 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(4^{x - 5}) = log₂(16)

log₂[(2²)^{x - 5}] = log₂(2⁴)

log₂(2^{2(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 :

3^{2x - 4} = 9^{3x + 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(3^{2x - 4}) = log₃(9^{3x + 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 :

16^{3m - 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(16^{3m - 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 :

3^x = 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(3^x) = ln(17)

x ⋅ ln(3) = ln(17)

Using calculator,

x ≈ 2.5789

5. Answer :

20^y = 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(20^y) = ln(56)

y ⋅ ln(20) = ln(56)

Using calculator,

y ≈ 1.3437

6. Answer :

5 ⋅ 18^6k = 26

Divide both sides by 5.

18^6k = 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(18^6k) = ln(5.2)

(6k)ln(18) = ln(5.2)

k ≈ 1.3437

7. Answer :

e^{x - 1} - 5 = 5

Add 5 to both sides.

e^{x - 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(e^{x - 1}) = ln(10)

(x - 1)ln(e) = ln(10)

(x - 1)(1) = ln(10)

x - 1 = ln(10)

x = ln(10) + 1

x ≈ 3.3026

8. Answer :

-e^{6 - 9p} + 5 = -48.4

Subtract 5 from both sides.

-e^{6 - 9p} = -53.4

Multiply both sides by -1.

e^{6 - 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(e^{6 - 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

p ≈ 0.2247

9. Answer :

8^{r + 4} ⋅ 4^{2r - 10} = 16^{r + 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(8^{r + 4} ⋅ 4^{2r - 10}) = log₂(16^{r + 1})

log₂(8^{r + 4}) + log₂(4^{2r - 10}) = log₂(16^{r + 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 :

A^{x + B} = C^Dx

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(A^{x + B}) = ln(C^Dx)

(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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