Question

SOlve for x (2)^2x-1 +5(2)^x =4

Answer 1

`{2}^(2x - 1) + 5 ({2}^(x) ) = 4`

`{2}^(2x) + 10( {2}^(x) ) = 8`

`( { {2}^(x) })^(2) + 10( {2}^(x)) + 25 = 33`

`{( {2}^(x) + 5)}^(2) = 33`

`{2}^(x) + 5 = √(33)`

`{2}^(x) = √(33) - 5`

`x = log_(2)( √(33) - 5 )`

Answer 2

To solve the equation (2)^(2x-1) + 5(2)^x = 4, we can first simplify it by combining like terms. Next, we can substitute u = 2^x, which gives us u^2 + 5u = 4. This is a quadratic equation that can be solved by factoring or using the quadratic formula.

Step-by-step explanation:

To solve the equation (2)^(2x-1) + 5(2)^x = 4, we can first simplify it by combining like terms:

We can rewrite (2)^(2x-1) as (2^x)^2, and rewrite 5(2)^x as 5 * 2^x. The equation now becomes:

(2^x)^2 + 5 * 2^x = 4

Next, we can substitute u = 2^x, which gives us:

u^2 + 5u = 4

This is a quadratic equation that can be solved by factoring or using the quadratic formula. Let's solve it by factoring:

u^2 + 5u - 4 = 0

(u + 4)(u - 1) = 0

Setting each factor equal to zero, we get:

u + 4 = 0 or u - 1 = 0

u = -4 or u = 1

Substituting back u = 2^x, we have:

2^x = -4 or 2^x = 1

Since 2^x can never be negative, the only possible solution is:

2^x = 1

Taking the logarithm of both sides, we get:

x = log2(1)

The logarithm of any number to the base 2 is 0, so the solution is:

x = 0