How do you solve the equation $\frac{1}{4}(8^{2x+1})+5\cdot8^x - 3 = 0$?

Question

$$ \frac{1}{4}(8^{2x+1})+5\cdot8^x - 3 = 0 $$

My work:

$t = 8^x$

$\frac{1}{4}(t^{2+1})+5t - 3 = 0$

Then I don’t know how to do it, but I think I did it wrong.

Answer 1

$\frac{1}{4}(t^2+1) +5t - 3 = 0$

$(t^2+1) + 20t - 12= 0$

$ t^2 + 20t - 11= 0$

Find it's roots using suitable method.

Here are some methods

And for this question you have to use quadratic formula.

Edit :

You have two different equation in your question. According to other solution is -

$\frac{1}{4}(8^{2x}\cdot8) +5\cdot8^x - 3 = 0$

$2\cdot8^{2x} +5\cdot8^x - 3 = 0$

Put $8^x = t$

$2t^2 + 5t - 3 = 0$

$2t^2 + 6t - 1t - 3 = 0$

$2t(t + 3) - 1(t + 3) = 0$

$(t + 3) (2t - 1) = 0$

$t = -3$ and $ t = \frac{1}{2}$

So $8^x = \frac{1}{2}$

$2^{3x} = 2^{-1}$

On equating powers,

$3x = -1$

$x = \frac{-1}{3}$

Answer 2

The title right now does not match the text, but presumably the equation to solve is

$${1\over4}8^{2x+1}+5\cdot8^x-3=0$$

The key here is that, when you let $t=8^x$, then $8^{2x+1}=8\cdot8^{2x}=8t^2$, not $t^{2+1}$. You get a quadratic,

$$2t^2+5t-3=(2t-1)(t+3)=0$$

the roots of which are $t=1/2$ and $t=-3$. The equation $8^x=1/2$, which can be rewritten as $2^{3x}=2^{-1}$, has solution $x=-1/3$. The equation $8^x=-3$ has no solution (in real numbers, at least). So $x=-1/3$ is the only solution.