SOLVING EXPONENTIAL EQUATIONS WITH QUADRATICS

Example 2 :

4^x – 10(2^x) + 16  =  0

Solution :

4^x – 10(2^x) + 16  =  0

(2²)^x – 10(2^x) + 16  =  0

(2^x)² – 10(2^x) + 16  =  0

Let u  =  2^x

u² – 10u + 16  =  0

Now, we have the quadratic equation form,

u² – 10u + 16  =  0

By factorization, we get

(u – 2) (u – 8)  =  0

u  =  2 and u  =  8

Replace u  =  2^x

So, 2^x  =  2 or 2^x  =  8

By using one to one property of exponential functions,

2^x  =  2 or 2^x  =  2³

x  =  1 or 3

So, the solution is {1 or 3}.

Example 3 :

4^x + 2^x  =  20

Solution :

4^x + 2^x  =  20

(2²)^x + 2^x  =  20

(2^x)² + 2^x - 20  =  0

Let u  =  2^x

u² + u - 20  =  0

By factorization, we get

(u – 4) (u + 5)  =  0

u  =  4 and u  =  -5

Replace u  =  2^x

So, 2^x  =  4 or 2^x  =  -5

We taking only positive.

By using the one to one property of exponential functions,

2^x  =  2²

x  =  2

So, the solution is 2.

Example 4 :

9^x + 3(3^x) + 2  =  0

Solution :

9^x + 3(3^x) + 2  =  0

 (3²)^x + 3(3^x) + 2  =  0

(3^x)² + 3(3^x) + 2  =  0

Let u  =  3^x

u² + 3u + 2  =  0

By factorization, we get

(u + 1) (u + 2)  =  0

u  =  -1 and u  =  -2

Replace u  =  2^x

So, 3^x  =  -1 or 3^x  =  -2

There are no values for x.

So, it has no solution.

Example 5 :

25^x + 4(5^x)  =  5

Solution :

25^x + 4(5^x)  =  5

(5²)^x + 4(5^x)  =  5

(5^x)² + 4(5^x) - 5  =  0

Let u  =  5^x

u² + 4u - 5  =  0

By factorization, we get

(u – 1) (u + 5)  =  0

u  =  1 and u  =  -5

Replace u  =  5^x

So, 5^x  =  1 or 5^x  =  -5

We taking only positive.

Since a⁰  =  1

5^x  =  1

x  =  0

So, the solution of x is 0

Example 6 :

25^x  =  23(5^x) + 50

Solution :

25^x  =  23(5^x) + 50

By combining like terms

 (5²)^x - 23(5^x) – 50  =  0

(5^x)² - 23(5^x) - 50  =  0

Let u  =  5^x

u² - 23u - 50  =  0

By factorization, we get

(u – 25) (u + 2)  =  0

u  =  25 and u  =  -2

Replace u  =  5^x

So, 5^x  =  25 or 5^x  =  -2

We taking only positive.

5^x  =  5²

x  =  2

So, the solution of x is 2

Example 7 :

49^x + 7  =  8(7^x)

Solution :

49^x + 7  =  8(7^x)

(7²)^x + 7  =  8(7^x)

(7^x)² + 7  =  8(7^x)

By combining like terms,

(7^x)² – 8(7^x) + 7  =  0

Let u  =  7^x

u² – 8u + 7  =  0

By factorization, we get

(u – 1) (u – 7)  =  0

u  =  1 and u  =  7

Replace u  =  7^x

So, 7^x  =  1 or 7^x  =  7

7^x  =  1 or 7^x  =  7

x  =  0 or 1

So, the solution is {0 or 1}.

Example 8 :

64^x + 8  =  6(8^x)

Solution :

64^x + 8  =  6(8^x)

(8²)^x + 8  =  6(8^x)

(8^x)² + 8  =  6(8^x)

By combining like terms,

(8^x)² – 6(8^x) + 8  =  0

Let u  =  8^x

u² – 6u + 8  =  0

By factorization, we get

(u – 2) (u – 4)  =  0

u  =  2 and u  =  4

Replace u  =  8^x

So, 8^x  =  2 or 8^x  =  4

(2³)^x  =  2 or (2³)^x  =  4

2^3x  =  2 or 2^3x  =  2²

By using the one to one property of exponential functions,

3x  =  1 or 3x  =  2

x  =  1/3 or x  =  2/3

So, the solution is {1/3 or 2/3}.

Example 9 :

If x and y are positive integers and 12³ = 2^x 3^y, what is the value of x + y ?

Solution :

12³ = 2^x 3^y

(2² ⋅ 3¹) = 2^x 3^y

x = 2 and y = 1

x + y = 2 + 1

= 3

So, the value of x + y is 3.

Example 10 :

If 2x 59nd y are positive integers and 12³ = 2^x 3^y, what is the value of x + y ?

Solution :

12³ = 2^x 3^y

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