# SOLVING EQUATIONS WITH EXPONENTS WORKSHEET

Solving Equations with Exponents Worksheet :

Worksheet given in this section will be much useful for the students who would like to practice problems on solving equations with exponents.

To solve equations with exponents, we need to be aware of laws of exponents.

To know more about laws of exponents,

## Solving Equations with Exponents Worksheet - Problems

Problem 1 :

Solve for k :

3x2 ⋅ 2x3  =  6xk

Problem 2 :

If 3x  =  10 and 3x-k  =  10/27, then solve for k.

Problem 3 :

If 2x+3 - 2x  =  k(2x), then solve for k.

Problem 4 :

If xac ⋅ xbc  =  x30, x > 1 and a + b  =  5, then solve for c.

Problem 5 :

If x2y3  =  10 and x3y2  =  8, then find the value of x5y5.

Problem 6 :

If x8y7  =  333 and x7y6  =  3, then find the value of xy.

Problem 7 :

If x2   =  y3 and x3z  =  y9, then solve for a.

Problem 8 :

If (√9)-7 ⋅ (√3)-4  =  3k, then solve for k.

Problem 9 :

If x√x  =  (x√x)x, then solve for x.

Problem 10 :

If 2x  =  3y  =  6-z, then find the value of

1/x + 1/y + 1/z ## Solving Equations with Exponents Worksheet - Solutions

Problem 1 :

Solve for k :

3x2 ⋅ 2x3  =  6xk

Solution :

3x2 ⋅ 2x3  =  6xk

(3 ⋅ 2)(x⋅ x3)  =  6xk

(6)(x2+3)  =  6xk

6x5  =  6xk

Divide each side by 6.

x5  =  xk

Using laws of exponents, we have

5  =  k

So, the value of k is 5.

Problem 2 :

If 3x  =  10 and 3x-k  =  10/27, then solve for k.

Solution :

3x-k  =  10/27

Using laws of exponents, we have

3x/3k  =  10/27

Substitute 10 for 3x.

10/3k  =  10/27

Take reciprocal on both sides.

3k/10  =  27/10

Multiply each side by 10.

3k  =  27

3k  =  33

Using laws of exponents, we have

k  =  3

So, the value of k is 3.

Problem 3 :

If 2x+3 - 2x  =  k(2x), then solve for k.

Solution :

Using laws of exponents, we have

2x ⋅ 23 - 2x =  k(2x)

2x ⋅ 8 - 2x =  k(2x)

2x(8 - 1)  =  k(2x)

2x(7)  =  k(2x)

Divide each side by 2x.

7  =  k

So, the value of k is 7.

Problem 4 :

If xac ⋅ xbc  =  x30, x > 1 and a + b  =  5, then solve for c.

Solution :

xac ⋅ xbc  =  x30

Using laws of exponents, we have

xac+bc  =  x30

ac + bc  =  30

Factor.

c(a + b)  =  30

Substitute 5 for (a + b).

5c  =  30

Divide each side by 5.

c  =  6

So, the value of c is 6.

Problem 5 :

If x2y3  =  10 and x3y2  =  8, then find the value of x5y5.

Solution :

x2y3  =  10 -----(1)

x3y2  =  8 -----(2)

Multiply (1) and (2) :

(1) ⋅ (2) -----> (x2y3) ⋅ (x3y2)  =  10 ⋅ 8

x5y5  =  80

So, the value x5yis 80.

Problem 6 :

If x8y7  =  333 and x7y6  =  3, then find the value of xy.

Solution :

x8y7  =  333 -----(1)

x7y6  =  3 -----(2)

Divide (1) by (2) :

(1) ÷ (2) -----> (x8y7) ÷ (x7y6)  =  333 ÷ 3

(x8y7) / (x7y6)  =  333 / 3

Simplify.

x8-7 ⋅ y7-6  =  111

x⋅ y1  =  111

xy  =  111

So, the value xy is 111.

Problem 7 :

If x2   =  y3 and x3z  =  y9, then solve for z.

Solution :

x3z  =  y9

x3z  =  y⋅ 3

x3z  =  (y3)3

Substitute xfor y3.

x3z  =  (x2)3

x3z  =  x6

3z  =  6

Divide each side by 3.

z  =  2

So, the value of z is 2.

Problem 8 :

If (√9)-7 ⋅ (√3)-4  =  3k, then solve for k.

Solution :

(91/2)-7 ⋅ (31/2)-4  =  3k

(9)-7/2 ⋅ (3)-4/2  =  3k

(32)-7/2 ⋅ 3-2  =  3k

3⋅ (-7/2) ⋅ 3-2  =  3k

3-7 ⋅ 3-2  =  3k

3-7 - 2  =  3k

3-9  =  3k

k  =  -9

So, the value of k is -9.

Problem 9 :

If x√x  =  (x√x)x, then solve for x.

Solution :

x√x  =  (x√x)x

If there is no exponent for any term, we can assume that the exponent of the term is 1.

So, we have

(x√x)1  =  (x√x)x

Using the properties of x, we have

1  =  x

So, the value of x is 1.

Problem 10 :

If 2x  =  3y  =  6-z, then find the value of

1/x + 1/y + 1/z

Solution :

Let 2x  =  3y  =  6-z  =  k.

Then we have

2x  =  k -----> 2  =  k1/x

3y  =  k -----> 3  =  k1/y

6-z  =  k -----> 6  =  k-1/z

Now, we have

⋅ 3  =  6

k1/x ⋅ k1/y  =  k-1/z

Using laws of exponents, we have

k1/x + 1/y  =  k-1/z

1/x + 1/y  =  -1/z

1/x + 1/y + 1/z  =  0

So, the value of 1/x + 1/y + 1/z is 0. After having gone through the stuff given above, we hope that the students would have understood, how to solve equations with exponents.

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