What is the positive value of x that satisfies the equation 2^{2x+1} - 9 \cdot 2^x + 4 = 0… — JAMB UTME

What is the positive value of $x$ that satisfies the equation $2^{2x+1} - 9 \cdot 2^x + 4 = 0$ ?

Answer: 2

The positive value of

x

that satisfies the equation is 2.

Applying the law of indices

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

gives

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

. Setting

y = 2^x

yields the quadratic equation

2y^2 - 9y + 4 = 0

. Factoring this expression gives

(2y - 1)(y - 4) = 0

, which yields roots

y = \frac{1}{2}

and

y = 4

. Solving

2^x = \frac{1}{2}

gives

x = -1

, and solving

2^x = 4

gives

x = 2

. The positive value is

2

Use index laws to express the equation in terms of

2^x

2 \cdot (2^x)^2 - 9 \cdot (2^x) + 4 = 0

By the product law of indices,

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

Substitute

y = 2^x

to form a quadratic equation

2y^2 - 9y + 4 = 0

Replacing

2^x

with a single variable simplifies the exponential equation into quadratic form.

Solve the quadratic equation for

y

y = \frac{1}{2}

y = 4

Factoring

2y^2 - 9y + 4 = 0

gives

(2y - 1)(y - 4) = 0

Substitute back

y = 2^x

to solve for

x

x = -1

x = 2

Since

2^x = \frac{1}{2} = 2^{-1}

x = -1

. Since

2^x = 4 = 2^2

x = 2

Select the positive value requested by the question

x = 2

x = 2

is positive, whereas

x = -1

is negative.

Reducing exponential equations to quadratic form using index laws

Estimated Time:1m 30s

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