SOLVING EQUATIONS WITH VARIABLE EXPONENTS

Example 1 :

If 3^a+1 = 3^-a+7, what is the value of a ?

Solution :

3^a+1 = 3^-a+7

On both sides of the equal sign, we have same base. So, equate the powers.

a+1 = -a+7

Add a and subtract 1 on both sides, we get

a+a = 7-1

2a = 6

a = 3

Example 2 :

If 3^x+2 = y, then what is the value of 3^x in terms of y ?

(a) y+9  (b) y-9  (c) y/3  (d)  y/9

Solution :

3^x+2 = y

Using the property a^m+n = a^m ⋅ aⁿ

3^x ⋅ 3² = y

3^x ⋅ 9 = y

Dividing by 9 on both sides, we get

3^x = y/9

Example 3 :

If 2a-b = 4, what is the value of 4^a/2^b

Solution :

Given :

2a-b = 4

From 4^a/2^b,

4 = 2 ⋅ 2 ==> 2²

4^a/2^b = (2²)^a/2^b

We have power raised to another power. So, multiply the powers.

= 2^2a/2^b

= 2^2a-b

By applying the value of 2a-b = 4, we will get

= 2⁴

= 16

Example 4 :

If 2^x+3 - 2^x = k(2^x), what is the value of k ?

Solution :

2^x+3 - 2^x = k(2^x)

Using the property a^m+n = a^m ⋅ aⁿ

2^x+3 - 2^x = k(2^x)

2^x ⋅ 2³- 2^x = k(2^x)

Factoring 2^x,

2^x (2³- 1) = k(2^x)

Divide by 2^x, we get

(2³- 1) = k

k = 8-1

k = 7

Example 5 :

If 

√(x√x) = x^a

then what is the value of a ?

Solution :

√(x√x) = x^a

We can write √x as x^1/2

(x√x)^1/2 = x^a

(x√x)^1/2 = x^a

Take squares on both sides, we get

(x√x) = (x^a)²

(x√x) = x^2a

(x ⋅ x^1/2) = x^2a

x^1+1/2 = x^2a

x^3/2 = x^2a

2a = 3/2

a = 3/4

Example 6 :

If n³ = x and n⁴ = 20x, where n > 0, what is the value of x ?

Solution :

n³ = x ----(1)

n⁴ = 20x ----(2)

Applying the value of x in (1), we get

n⁴ = 20n³

n = 20

By applying the value of x in (1), we get

x = 20³

x = 8000

Example 7 :

Solve for x, 2(4^2x+1) = 128

Solution :

2(4^2x+1) = 128

Divide by 2 on both sides.

4^2x+1 = 64

64 = 4 ⋅ 4 ⋅ 4 ==>  4³

4^2x+1 = 4³

Bases are equal, so equate the powers.

2x+1 = 3

2x = 2

x = 1

Example 8 :

12^2x-1 = (∜12)^x

Solution :

12^2x-1 = (∜12)^x

12^2x-1 = ((12)^1/4)^x

12^2x-1 = (12)^x/4

2x-1 = x/4

4(2x-1) = x

8x-4 = x

8x-x = 4

x = 4/7

Example 9 :

3^4x+3 = 81^x

Solution :

3^4x+3 = 81^x

81 = 3⁴

3^4x+3 = (3⁴)^x

3^4x+3 = 3^4x

4x+3 = 4x

So, there is no solution.

Example 10 :

12 ⋅ 2^x-7 = 24

Solution :

12 ⋅ 2^x-7 = 24

Divide by 12 on both sides, we get

2^x-7 = 2¹

x-7 = 1

x = 1+7

x = 8

Example 11 :

If (-a² b³) ⋅ (2a b²) (-3b) = ka^m bⁿ

What is the value of m + n ?

Solution :

(-a² b³) ⋅ (2a b²) (-3b) = ka^m bⁿ

6 a²⁺¹ b^{3 + 2 + 1} = ka^m bⁿ

6 a³ b⁶ = ka^m bⁿ

Comparing the corresponding terms, we get

k = 6, m = 3 and n = 6

m + n = 3 + 6

= 9

So, the value of m + n is 9.

Example 12 :

If (2/3 a² b)² ⋅ (4/3 ab)^-3 = ka^m bⁿ

What is the value of k ?

Solution :

(2/3 a² b)² ⋅ (4/3 ab)^-3 = ka^m bⁿ

(2/3)²  (a²)²  b²⋅ (4/3)^-3 (ab)^-3 = ka^m bⁿ

(4/9)²  a⁴ b²⋅ (3/4)³ 1/(ab)³ = ka^m bⁿ

(16/81) ⋅ (27/64) a⁴ b² (1/a³b³) = ka^m bⁿ

(1/3) ⋅ (1/4) (a⁴ b² / a³b³) = ka^m bⁿ

(1/12) (a/b) = ka^m bⁿ

(1/12) (ab^-1) = ka^m bⁿ

Comparing the corresponding terms, k = 1/12, m = 1 and b = -1

So, the value of k is 1/12.

Example 13 :

If (x³) (-y)² z^-2 /(x)^-2 y³ z = x^m / yⁿ z^p

What is the value of m + n + p ?

Solution :

(x³) (-y)² z^-2 /(x)^-2 y³ z = x^m / yⁿ z^p

(x³) (x)²y²  / y³ z z² = x^m / yⁿ z^p

x^{3 + 2} / y^{3 - 2} z¹⁺² = x^m / yⁿ z^p

x⁵ / y z³ = x^m / yⁿ z^p

x⁵ / y z³ = x^m / yⁿ z^p

Comparing the corresponding terms, we get 

m = 5, n = 1 and p = 3

m + n + p = 5 + 1 + 3

= 9

So, the value of m + n + p is 9.

Example 14 :

If 3^(a + b)² / 3^(a - b)² = 243, what is the value of ab ?

a)  5/4   b)  3/2     c)  7/4     d)  2

Solution :

3^(a + b)² / 3^(a - b)² = 243

Using quotient rules in exponents, simplifying it 

3^(a + b)² -  (a - b)² = 243

343 = 7 x 7 x 7 ==> 3⁵

3^(a² + 2ab + b²) - (a² - 2ab + b²) = 3⁵

3^(a² + 2ab + b² - a² + 2ab - b²) = 3⁵

3^(4ab) = 3⁵

3^4ab = 3⁵

4ab = 5

ab = 5/4

So, option a is correct.