# Math Help - Solve 4(2^x)+31(2^x)-8=0

1. ## Solve 4(2^x)+31(2^x)-8=0

HEY GUYS,

I'm working my way through an online course and it is impossible getting a phone call back from my teachers. This equation has been bothering me for days now and its holding me back.

4(2^x)+31(2^x)-8=0

The lesson i'm on is exponential and logarithmic equations, and i understand i'm supposed to find common bases' and apply logarithms, but for some reason this question is beyond me. Any help in the right direction would be much appreciated. THANKS

2. You have like terms, so
$4(2^x)+31(2^x)-8=0$ becomes

$35(2^x)-8=0$

$35(2^x) = 8$

divide by 35

$2^x = 8/35$

natural logs, etc...

$x = \frac{ln{(8/35)}}{ln(2)}$

-----------------------------------

Now, I suspect that there is a typo in your original problem. If you confirm this, then I will work out the quadratic.

3. Originally Posted by INEEDHELLP
HEY GUYS,

I'm working my way through an online course and it is impossible getting a phone call back from my teachers. This equation has been bothering me for days now and its holding me back.

4(2^x)+31(2^x)-8=0

The lesson i'm on is exponential and logarithmic equations, and i understand i'm supposed to find common bases' and apply logarithms, but for some reason this question is beyond me. Any help in the right direction would be much appreciated. THANKS
Is there a typo here?
$\displaystyle 4 \cdot 2^x + 31 \cdot 2^x - 8 = 0$

Collect like terms:
$\displaystyle 35 \cdot 2^x = 8$

$\displaystyle 2^x = \frac{8}{35}$

Take logs (base 2):
$\displaystyle x = log_2 \left ( \frac{8}{35} \right )$

or, if you prefer:
$\displaystyle x = log_2(8) - log_2(35)$

-Dan

TheChaz is too fast for me!

4. Originally Posted by topsquark
Is there a typo here?...
TheChaz is too fast for me!
I'll make you a deal: if there turns out to indeed be a typo, I'll let you handle it! Unless, of course, our abacus friend makes an appearance! He's pretty quick on the draw (but also a perfectionist, which can cost precious seconds )

5. Much appreciate guys, i know it seems simple but when you crunching numbers all day, the simplest things become the hardest. You guys are awesome for the quick responses. Thank you.

6. No problem.

But answer me this: are you sure that you have copied (and we have worked) the correct problem?? Notice that two of us separately suspected a typo. There's good reason, too!

7. Yep, the equation I'm given is

4(2^x) + 31(2^x) - 8 = 0

Thanks again.

8. Hello, INEEDHELLP!

From the "shape" of the equation, I suspect that there IS a typo.

I believe the equation is: . $4\!\cdot\!2^{2x}+31\!\cdot\!2^x -8\:=\:0$
. . This would explain the unusual coefficients.

Factor: . $\left(2^x + 8\right)\left(4\!\cdot\!2^x - 1\right) \:=\:0$

And we have:

. . $2^x + 8 \:=\:0 \quad\Rightarrow\quad 2^x \:=\:-8\quad\text{No real roots}$

. . $4\!\cdot\!2^x - 1 \:=\:0 \quad\Rightarrow\quad 2^x \:=\:\frac{1}{4} \quad\Rightarrow\quad 2^x \:=\:2^{-2} \quad\Rightarrow\quad \boxed{x \:=\:-2}$

10. sorry guys, but there isn't a typo, that is all of it.

4(2^x) + 31(2^x) - 8 = 0

Just out of curiosty, how come everyone thinks its a typo? Tell me, i want to learn!

11. When you start with three terms, all on the same side of the equation.
When the middle coefficient (31) is the sum of two factors of the product of the other coefficients ("ac" = 4*-8 = -32 = 32 * -1; 32 + -1 = 31 = "b").

When your life is no longer your own. Oh wait, that's something else.

12. Originally Posted by INEEDHELLP

Just out of curiosty, how come everyone thinks its a typo? Tell me, i want to learn!
The form of equation (which leads to a quadratic) you posted is a similar theme here on this forum so Soroban and others assumed it was again a similar question. The version you initially posted that topsquark solved is a lot easier...

13. Ya, the whole prime of 31 really did a number on my brain. I just couldn't rap my head around it.