# SOLVING EXPONENTIAL EQUATIONS WORKSHEET

Problems 1-5 : Solve for x.

Problem 1 :

3x = 27

Problem 2 :

2x - 5 = ¹⁄₁₆

Problem 3 :

⋅ 3x = 54

Problem 4 :

4= 7(2x) + 18

Problem 5 :

5= (√25)-7 ⋅ (√5)-5)

Problems 6-10 : Solve for x using logarithm.

Problem 6 :

10x = 100

Problem 7 :

3x = ¹⁄₂₇

Problem 8 :

53x - 1 = ¹⁄₆₂₅

Problem 9 :

2x - 1 = 5

Problem 10 :

5x - 1 - 2x = 0

3x = 27

3x = 3 ⋅ 3 ⋅ 3

3x = 33

Equate the exponents.

x = 3

2x - 5 = 2-4

Equate the exponents.

x - 5 = -4

x = 1

⋅ 3x = 63

Divide both sides by 7.

3x = 9

3x = 32

Equate the exponents.

x = 2

4= 7(2x) + 18

(22)= 7(2x) + 18

(2x)= 7(2x) + 18

(2x)- 7(2x) - 18 = 0

Let a = 2x.

a- 7a - 18 = 0

(a - 9)(a + 2) = 0

a - 9 = 0  or  a + 2 = 0

 a - 9 = 0a = 9a = 323x = 32x = 2 a + 2 = 0a = -2a = -23x = -2

In 3x, whatever real value (positive or negative or zero) we substitute for x, 3x can never be negative. So we can ignore the equation 3x = -2.

Therefore,

x = 2

5= (√25)-7 ⋅ (√5)-5

5= 5-7 ⋅ (51/2)-5

5= 5-7 5-5/2

5= 5-7 - 5/2

5= 5-19/2

Equate the exponents.

x = -19/2

10x = 100

The above equation is in exponential form. Convert it to logarithmic form to solve for x.

x = log10100

x = log10100

x = log10(102)

x = 2log1010

x = 2(1)

x = 2

3x = ¹⁄₂₇

The above equation is in exponential form. Convert it to logarithmic form to solve for x.

x = log3(¹⁄₂₇)

x = log31 - log327

x = 0 - log3(33)

x = -3log33

x = -3(1)

x = -3

53x - 1 = ¹⁄₆₂₅

The above equation is in exponential form. Convert it to logarithmic form to solve for x.

3x - 1 = log5(¹⁄₆₂₅)

3x - 1 = log51 - log5625

3x - 1 = 0 - log5(54)

3x - 1 = -4log55

3x - 1 = -4(1)

3x - 1 = -4

3x = -3

Divide both sides by 3.

x = -1

2x - 1 = 5

2x = 5

On the left side of the equation above, the base is 2. On the right side, 6 is not a power of 2. So, we can not get the same base on both sides.

Take logarithm on both sides and solve for x.

log(2x) = log6

Using power rule of logarithm,

xlog2 = log6

Divide both sides by log2.

5x - 1 - 2x = 0

5x - 1 = 2x

In the equation above, the base on the right side and the base left side are not same. And also, we can not make the base same on both sides using the rules of exponents.

Take logarithm on both sides and solve for x.

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

Using power rule of logarithm,

(x - 1)log5 = xlog2

Using distributive property,

xlog5 - log5 = xlog2

Subtract xlog2 from both sides.

xlog5 - log5 - xlog2 = 0

xlog5 - xlog2 = log5

Factor.

x(log5 - log)2 = log5

Divide both sides by (log5 - log2).

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