Question

Question #b397b

Answer 1

`x = 1/2`

Explanation:

Start by rewriting your equation

`2^(2x) - 5 * 4^(x+1) + 38 = 0`

as

`2^(2x) - 5 * 4^x * 4 + 38 = 0`

As you know, you have

`4 = 2^2`

This implies that

`4^x = (2^2)^x = 2^(2 * x) = 2^(2x)`

This means that the equation can be written as

`2^(2x) - 5* 4 * 2^(2x) + 38 = 0`

At this point, you can take `2^(2x)` as a common factor and say that

`2^(2x) * (1 - 5 * 4) + 38 = 0`

This is equivalent to

`2^(2x) = (- 38)/(-19)`

`2^(2x) = 2`

Since `2` is simply `2^1`, you can say that

`2^(2x) = 2^1`

This implies that

`2x = 1`

which gets you

`x = 1/2`

To double-check your calculations, plug `x = 1/2` into the original equation.

`2^((2 * 1/2)) - 5 * 4^((1/2 + 1)) + 38 = 0`

`2^1 - 5 * 4^(3/2) + 38 = 0`

Since

`4^(3/2) = sqrt(4^3) = 4sqrt(4) = 4 * 2 = 8`

you will have

`2 - 5 * 8 + 38 = 0`

`2 - 40 + 38 = 0 " "color(darkgreen)(sqrt())`

Answer 2

`x=1/2`

Explanation:

Note first that `5(4^(x+1))=5(4^x)4^1=20(4^x)`.

`2^(2x)-5(4^(x+1))+38=0`

`2^(2x)-20(4^x)+38=0`

Rewrite the exponential function with base `4` as one with base `2` so that we are working with a standard base throughout. Note that `4^x=(2^2)^x=2^(2x)`.

`2^(2x)-20(2^(2x))+38=0`

Now, note that `2^(2x)-20(2^(2x))=-19(2^(2x))`. This is just like how `x-20x=-19x`.

`-19(2^(2x))+38=0`

To solve for `x`, first isolate `2^(2x)`.

`-19(2^(2x))=-38`

Dividing by `-19`:

`2^(2x)=2`

`2^color(blue)(2x)=2^color(blue)1`

Since the bases of the exponential functions are equal, so must their exponents:

`2x=1`

`x=1/2`