SOLVING EQUATIONS WITH EXPONENTS WORKSHEET

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

Detailed Answer Key

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

Add 1/z to each side. 

1/x + 1/y + 1/z  =  0

So, the value of 1/x + 1/y + 1/z is 0.

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