SOLVING EXPONENTIAL EQUATIONS

Problem 1 :

If a and b are positive integers and 5^{a + b} = 5^3c, which of the following is equal to 5^9c ?

a)  5^{3a + b}      b) 15^{a + b}        c) 25^{3a + 3b}        d)  125^{a + b}     e) 125^{3a + 3b}

Solution :

Given that, 5^{a + b} = 5^3c

To find the value of 5^9c, we raise power 3 on both sides.

(5^{a + b})³ = (5^3c)³

5^{3(a + b)} = 5^9c

(5³)^{(a + b)} = 5^9c

The value of 5³ is 5 x 5 x 5 which is 125.

125^{(a + b)} = 5^9c

Option d is correct.

Problem 2 :

(x² y³)^1/2 (x² y³)^1/3 = x^a/3  y^a/2

If the equation above, where a is constant is true for all positive values of x and y, what is the value of a ?

a)  2      b)  3      c)  5      d)  6

Solution :

(x² y³)^1/2 (x² y³)^1/3 = x^a/3  y^a/2

(x^2/2 y^3/2) (x^2/3 y^3/3) = x^a/3  y^a/2

(x y^3/2) (x^2/3 y) = x^a/3  y^a/2

Combining the powers, we get

x^{(1 + 2/3)} y ^{(3/2 + 1)} = x^a/3  y^a/2

x^5/3 y^5/2 = x^a/3  y^a/2

Comparing the powers of corresponding terms, we get

In both ways, we should get the same answer. So, the value of a is 5.

Problem 3 :

If 49 = (7y)² then y could equal which of the following 

I   -1       II     1     III   7

a)  II only     b)  III only      c)  I and II      d)  I and III      d)  II and III

Solution :

Given that 49 = (7y)² 

49 = 7²y² 

49 = 49y² 

Dividing by 49 on both sides

49/49 = y²

y² = 1

y = ± 1

So, the values of y are -1 and 1. Option c is correct.

Problem 4 :

If (a⁶ a⁶)^k = (a³)^{k + 5} , where a is a constant greater than 1, what is the value of k ?

a)  3/2      b)  5/3     c)  2       d)  3     e)  5

Solution :

(a⁶ a⁶)^k = (a³)^{k + 5}

(a⁶⁺⁶)^k = a^{3(k + 5)}

a^12k = a^{3(k + 5)}

Since the bases are equal, then we can equate the powers.

12k = 3(k + 5)

12k = 3k + 15

12k - 3k = 15

9k = 15

k = 15/9

k = 5/3

So, the value of k is 5/3, which is option b is correct.

Problem 5 :

If t and w are positive numbers and t⁸ / t² = w¹², then w⁸ / w²

a)  t²      b)  t³       c)  t⁴        d)  t⁶       e)  t¹²

Solution :

Given that, t⁸ / t² = w¹²

t^8-2 = w¹²

t⁶ = w¹² ----(1)

Simplifying that we have, w⁸ / w², w^8-2 = w⁶

From (1), t⁶ = w¹²

 w¹² =  t⁶

 (w⁶)² =  t⁶

Moving power 2 to the other side of equal sign, we get

w⁶ = (t⁶)^1/2

= t^{6 x 1/2}

= t³

So, option b is correct.

Problem 6 :

If t > 0 and t² − 4 = 0 , what is the value of t ?

Solution :

t² − 4 = 0

t² = 4

t² = 2²

t = ±2

The value of t are -2 and 2. Since the given condition is t > 0, then the value of t is 2.

Problem 7 :

If a²  = 4, then a⁶ = 

a)  12      b)  16       c)  24        d)  32       e)  64

Solution :

a²  = 4, then a⁶ = 

Raising power 3 on both sides, we get

(a²)³ = 4³

a⁶ = 4 x 4 x 4

a⁶ = 64

So, the value of a⁶ is 64, option e is correct.

Problem 8 :

k^ (x² + xy) k^ (y² + xy) = k²⁵

In the equation above, k > 1 and x = 3 what is the positive value of y ?

a)  1      b)  2       c)  4        d)  5

Solution :

k^ (x² + xy) k^ (y² + xy) = k²⁵

Using the rule in exponents,

k^{(x^2 + xy) + (y^2 + xy)} = k²⁵

k^{(x^2 + xy + y^2 + xy)} = k²⁵

k^{(x^2 + 2xy + y^2)} = k²⁵

k^{(x + y)^2} = k²⁵

By equating the powers, we get

(x + y)² = 25

By applying the value of x as 3, we get

(3 + y)² = 25

3 + y = √25

3 + y = 5 (since we apply the positive value for y)

y = 5 - 3

y = 2

So, option b is correct.

Problem 9 :

If 3x - y = 12, what is the value of 8^x/2^y ?

a)  2¹²      b)  44       c)  82 

d) The value cannot be determined from the information given

Solution :

8^x/2^y = (2³)^x/2^y

= 2^3x/2^y

= 2^{3x - y}

Applying the value 3x - y = 12 above, we get

= 2¹²

Option a is correct.

Problem 10 :

25^x/5^x+3 = (1/125)^{x + 1}

Solution :

25^x/5^x+3 = (1/125)^{x + 1}

(5²)^x/5^x+3 = (1/5³)^{x + 1}

5^2x/5^x+3 = (5^-3)^{x + 1}

5^2x 5^-(x+3) = 5^{-3(x + 1)}

5^{2x - (x + 3)} = 5^{-3(x + 1)}

5^{2x - x - 3} = 5^{-3x - 3}

5^{x - 3} = 5^{-3x - 3}

x  - 3 = -3x - 3

x + 3x = -3 + 3

4x = 0

x = 0

So, the value of x is 0.