Find the product of all real solutions to the exponential equation 9^{x+1} - 28 \cdot 3^x … — JAMB UTME

Find the product of all real solutions to the exponential equation $9^{x+1} - 28 \cdot 3^x + 3 = 0$ .

Cevap: -2

The product of all real solutions is -2.

Using the index laws

9^{x+1} = 9 \cdot (3^x)^2

, let

u = 3^x

. The equation reduces to the quadratic

9u^2 - 28u + 3 = 0

, which factors as

(9u - 1)(u - 3) = 0

. This gives

u = 1/9

u = 3

. Solving

3^x = 1/9

yields

x = -2

, and solving

3^x = 3

yields

x = 1

. The product of these solutions is

(-2) \times 1 = -2

Express

9^{x+1}

in terms of

3^x

9^{x+1} = 9^1 \cdot 9^x = 9 \cdot (3^2)^x = 9 \cdot (3^x)^2

Apply index laws

a^{m+n} = a^m \cdot a^n

and

(a^m)^n = (a^n)^m

to establish a common base of 3.

Substitute

u = 3^x

to form a quadratic equation

9u^2 - 28u + 3 = 0

Transform the exponential equation into a standard quadratic algebraic equation.

Solve the quadratic equation for

u

(9u - 1)(u - 3) = 0 \implies u = \frac{1}{9}

u = 3

Factorize the quadratic expression to determine its roots.

Substitute back

u = 3^x

to solve for

x

3^x = 3^{-2} \implies x = -2

, and

3^x = 3^1 \implies x = 1

Equate exponents with matching bases to find all valid real solutions for

x

Find the product of the two solutions

(-2) \times 1 = -2

Calculate the required mathematical product of the solutions.

Solving exponential equations reducible to quadratic form using laws of indices

Soru