# How would I factor 4^2x + 2(4^x) - 3 = 0?

### 9 Answers

- PinkgreenLv 71 month ago
4^(2x)+2(4^x)-3=0

=>

(4^x)^2+2(4^x)-3=0

=>

[(4^x)+3][(4^x)-1]=0

or

x[x-ln(-3)/ln(4)]=0

- lenpol7Lv 71 month ago
4^2x + 2(4^x) - 3 = 0

(4^)^2 + 2(4^x) - 3 = 0

Think of 4^x = y

Substitute

y^2 + 2y - 3 = 0

Factor

( y + 3)(y - 1) = 0

Hence

4^x = -3 (Unresolved)

&

4^x = 1

x = 0

- la consoleLv 71 month ago
4^(2x) + [2 * 4^(x)] - 3 = 0

4^(x + x) + [2 * 4^(x)] - 3 = 0

[4^(x) * 4^(x)] + [2 * 4^(x)] - 3 = 0 → let: a = 4^(x) → where: a > 0 because exponential

[a * a] + [2 * a] - 3 = 0

a² + 2a - 3 = 0

a² + 2a = 3

a² + 2a + 1 = 4

(a + 1)² = 4

a + 1 = ± 2

a = - 1 ± 2 → recall the condition: a > 0

a = - 1 + 2

a = 1 → recall: a = 4^(x)

4^(x) = 1

x = 0

- Jeff AaronLv 71 month ago
(4^2)x + 2(4^x) - 3 = 0

Krishnamurthy gave the correct solution which uses the Product Log (Lambert) Function.

Or did you mean:

4^(2x) + 2(4^x) - 3 = 0

Let u = 4^x, so u^2 = 4^(2x), so we have:

u^2 + 2u - 3 = 0

u^2 + 3u - u - 3 = 0

u(u + 3) - (u + 3) = 0

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

u - 1 = 0 or u + 3 = 0

4^x = 1 or 4^x = -3

If 4^x = 1 then:

x = (ln(1) + 2*i*pi*n) / ln(4), for any integer n

x = (i*pi*n) / ln(2), for any integer n

x =~ 4.5323601418271938096276829457167in, for any integer n

If x is a real number, then n = 0, so x = 0

If 4^x = -3 then:

x = (ln(-3) + 2*i*pi*n) / ln(4), for any integer n

x = (ln(-3) + i*pi*n) / ln(4), for any even number n

x = (ln(3) + i*pi*n) / ln(4), for any odd number n

x =~ 0.79248125036057809072686947197391 + 2.2661800709135969048138414728583n, for any odd number n

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- rotchmLv 71 month ago
The idea here is to notice that your equation is a quadratic. Do you see it?

And since you know how to solve a quadratic, you are essentially done.

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.

- Iggy RockoLv 71 month ago
Assuming 4^2x should be 4^(2x),

4^(2x) + 2(4^x) - 3 = 0

Let u = 4^x.

u^2 - 2u - 3 = 0

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

u = 3 or u = -1

4^x = 3 or 4^x = -1

4^x = -1 has no real solutions, leaving just

4^x = 3

x = log(base 4)3

- Pramod KumarLv 71 month ago
This is an equation. If you wish to factorize the left side then proceed as follows :

4^(2x) + 2 ( 4^x ) - 3

This can be re-written as = (4^x)² + 2 ( 4^x ) - 3

Let 4^(x) = A then --

=> 4^(2x) + 2 ( 4^x ) - 3 = A² + 3 A - A - 3

=> A ( A + 3 ) - 1 ( A + 3 )

=> ( A - 1 ) ( A + 3 )

Putting A = 4^x

4^(2x) + 2 ( 4^2 ) - 3 = ( 4^x - 1 ) ( 4^x + 3 ) ................... Required factors.(Answer)

If you want to calculate x from this equation, then put --

( 4^x - 1 ) ( 4^x + 3 ) = 0

=> ( 4^x - 1 ) = 0 ===>>> 4^x = 1 ====>>> x = 0

And ( 4^x + 3 ) = 0 => 4^x = - 3

=> 2^(2x) = - 3

Power on the Left hand side is even, hence this equation will have two complex roots.

- Wayne DeguManLv 71 month ago
4²ˣ = (4ˣ)²

so, (4ˣ)² + 2(4ˣ) - 3 = 0

If y = 4ˣ then,

y² + 2y - 3 = 0

i.e. (y + 3)(y - 1) = 0

Hence, y = -3 or y = 1

so, 4ˣ = 1 or 4ˣ = -3...not possible

Then, x = 0 is the only solution

Checking gives: 4⁰ + 2(4⁰) - 3 => 1 + 2 - 3 = 0

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