Algebra.

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Hi, again

I've got 2 tasks.

$$2^{x+3}-2^{2-x}+31=0\;\;\;\;,\;\;\;x=?$

$$A)\;-1$

$$B)\;-3$

$$C)\;0$

$$D)\;1$

$\text{and:}$

$$2^x=50\;\;\;,\;\;\;\;2^{2x-2}=?$

$$A)\;625$

$$B)\;250$

$$C)\;500$

$$D)\;650$

Shadow · **Posted:** Wed, 11 Jan 2012 02:33:11 UTC

kreshnik wrote:

Hi, again

I've got 2 tasks.

$$2^{x+3}-2^{2-x}+31=0\;\;\;\;,\;\;\;x=?$

$$A)\;-1$

$$B)\;-3$

$$C)\;0$

$$D)\;1$

$\text{and:}$

$$2^x=50\;\;\;,\;\;\;\;2^{2x-2}=?$

$$A)\;625$

$$B)\;250$

$$C)\;500$

$$D)\;650$

For the first one, multiply through by 2^x

, and then move the rest of the stuff to the right side and solve. For the second one, square your given, and divide by 4.

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Quote:

For the first one, multiply through by 2^x

, and then move the rest of the stuff to the right side and solve. For the second one, square your given, and divide by 4.

I still can't solve it!

$$2^{x+3}-2^{2-x}+31=0\;\;,\;\;x=?$

$$2^{x+3}-2^{2-x}+31=0\;\;\;\;/*2^x$

$$2^x(2^{x+3}-2^{2-x}+31)=0*2^x$

$$2^{x+3+x}-2^{2-x+x}+31*2^{x}=0$

$$2^{2x+3}-2^2+31*2^x=0$

$$2^{2x+3}+31*2^x=2^2$

$$2^x(2^{x+3}+31)=2^2$

$$\text{I know that}\;\;\; x=-3\;\; \text{because:}$

$$2^{-3}(2^{-3+3}+31)=4$

$$\frac{1}{2^3}(2^0+31)=4$

$$\frac{32}{8}=4$

$4=4

But I need steps.
And I'm still stuck here. And for the second one, I have no idea what should I do, If you don't mind please try to help me with numbers not with words,
it's quite hard to understand you (because English is foreign language for me), if you're already bored with my dizzy tasks, just let me know, it's OK. :confused:

outermeasure · **Posted:** Wed, 11 Jan 2012 16:49:39 UTC

kreshnik wrote:

Quote:

For the first one, multiply through by 2^x

, and then move the rest of the stuff to the right side and solve. For the second one, square your given, and divide by 4.

I still can't solve it!

$$2^{x+3}-2^{2-x}+31=0\;\;,\;\;x=?$

$$2^{x+3}-2^{2-x}+31=0\;\;\;\;/*2^x$

$$2^x(2^{x+3}-2^{2-x}+31)=0*2^x$

$$2^{x+3+x}-2^{2-x+x}+31*2^{x}=0$

$$2^{2x+3}-2^2+31*2^x=0$

$$2^{2x+3}+31*2^x=2^2$

$$2^x(2^{x+3}+31)=2^2$

$$\text{I know that}\;\;\; x=-3\;\; \text{because:}$

$$2^{-3}(2^{-3+3}+31)=4$

$$\frac{1}{2^3}(2^0+31)=4$

$$\frac{32}{8}=4$

$4=4

But I need steps.
And I'm still stuck here. And for the second one, I have no idea what should I do, If you don't mind please try to help me with numbers not with words,
it's quite hard to understand you (because English is foreign language for me), if you're already bored with my dizzy tasks, just let me know, it's OK. :confused:

For the first one, you have a quadratic equation in 2^x

(namely

) which you should solve. What do you get for the second one following Shadow's suggestion?

Denis · **Posted:** Wed, 11 Jan 2012 18:19:46 UTC

On the 1st one, OM's suggestion will lead to 2^x = 1/8 or -4;
you will use 2^x = 1/8, NOT 2^x = -4 ; OK?

Also, usually easier if you substitute, like let a = 2^x:
8a^2 + 31a - 4 = 0

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

outermeasure wrote:

For the first one, you have a quadratic equation in 2^x

(namely

) which you should solve. What do you get for the second one following Shadow's suggestion?

Denis wrote:

On the 1st one, OM's suggestion will lead to 2^x = 1/8 or -4;
you will use 2^x = 1/8, NOT 2^x = -4 ; OK?

Also, usually easier if you substitute, like let a = 2^x:
8a^2 + 31a - 4 = 0

Thanks a lot guys I used a quadratic equation as outermeasure said, and I used Denis' advice: $2^x=a

and I did it like:

$8a^2+31a-4=0

$$a=8\;\;,\;\;b=31\;\;,\;\;c=-4$

$$D=b^2-4ac\;\;=>\;\;D=31^2-4*8*(-4)$

$$D=961+128\;\;=>\;\;D=1089$

$$x_1=\frac{-31-\sqrt{1089}}{2*8}$

$$x_1=\frac{-31-33}{16}\;\;=>\;\;x_1=\frac{-64}{16}\;\;=>\;\;x_1=-4$

For $x=-4

we have: $$2^x=-4\;\;=>\;\;2^x=-2^2\;\;\text{we can't use this!}$

$$x_2=\frac{-31+\sqrt{1089}}{2*8}$

$$x_2=\frac{-31+33}{16}\;\;=>\;\;x_2=\frac{2}{16}\;\;=>\;\;x_2=\frac{1}{8}$

For $$x=\frac18$ we have $$2^x=\frac{1}{8}\;\;=>\;\;2^x=2^{-3}\;\;=>\;\;x=-3$

For the second one... I've come up with a solution:

$$2^x=50\;\;\;,\;\;\;2^{2x-2}=?$

$2^x=25*2

$$\frac{2^x}{2}=25$

$$2^{x-1}=25\;\;\;\;/*2^{x-1}$

$2^{x-1}*(2^{x-1})=2^{x-1}*25$

$$2^{x-1+(x-1)}=2^{x-1}*25$

$$2^{2x-2}=2^{x-1}*25$

So we have now:

$$2^{2x-2}=2^{x-1}*25$

$$2^{2x-2}=\frac{2^x}{2}*25$

since: $$2^x=50\;\;\text{we have:}$

$$2^{2x-2}=\frac{50}{2}*25$

$$2^{2x-2}=25*25$

$$2^{2x-2}=625$

Thanks a lot again everyone for the help. :wink:

Now I would like to know how did shadow think about his way of doing it.

Shadow · **Posted:** Wed, 11 Jan 2012 22:15:58 UTC

kreshnik wrote:

outermeasure wrote:

For the first one, you have a quadratic equation in 2^x

(namely

) which you should solve. What do you get for the second one following Shadow's suggestion?

Denis wrote:

On the 1st one, OM's suggestion will lead to 2^x = 1/8 or -4;
you will use 2^x = 1/8, NOT 2^x = -4 ; OK?

Also, usually easier if you substitute, like let a = 2^x:
8a^2 + 31a - 4 = 0

Thanks a lot guys I used a quadratic equation as outermeasure said, and I used Denis' advice: $2^x=a

and I did it like:

$8a^2+31a-4=0

$$a=8\;\;,\;\;b=31\;\;,\;\;c=-4$

$$D=b^2-4ac\;\;=>\;\;D=31^2-4*8*(-4)$

$$D=961+128\;\;=>\;\;D=1089$

$$x_1=\frac{-31-\sqrt{1089}}{2*8}$

$$x_1=\frac{-31-33}{16}\;\;=>\;\;x_1=\frac{-64}{16}\;\;=>\;\;x_1=-4$

For $x=-4

we have: $$2^x=-4\;\;=>\;\;2^x=-2^2\;\;\text{we can't use this!}$

$$x_2=\frac{-31+\sqrt{1089}}{2*8}$

$$x_2=\frac{-31+33}{16}\;\;=>\;\;x_2=\frac{2}{16}\;\;=>\;\;x_2=\frac{1}{8}$

For $$x=\frac18$ we have $$2^x=\frac{1}{8}\;\;=>\;\;2^x=2^{-3}\;\;=>\;\;x=-3$

For the second one... I've come up with a solution:

$$2^x=50\;\;\;,\;\;\;2^{2x-2}=?$

$2^x=25*2

$$\frac{2^x}{2}=25$

$$2^{x-1}=25\;\;\;\;/*2^{x-1}$

$2^{x-1}*(2^{x-1})=2^{x-1}*25$

$$2^{x-1+(x-1)}=2^{x-1}*25$

$$2^{2x-2}=2^{x-1}*25$

So we have now:

$$2^{2x-2}=2^{x-1}*25$

$$2^{2x-2}=\frac{2^x}{2}*25$

since: $$2^x=50\;\;\text{we have:}$

$$2^{2x-2}=\frac{50}{2}*25$

$$2^{2x-2}=25*25$

$$2^{2x-2}=625$

Thanks a lot again everyone for the help. :wink:

Now I would like to know how did shadow think about his way of doing it.

$2^x = 50 \iff 2^{2x}=(2^x)^2=50^2=2500\quad 2^{2x-2}={1\over 4} 2^{2x} =625$

Denis · **Posted:** Thu, 12 Jan 2012 06:26:17 UTC

kreshnik wrote:

Now I would like to know how did shadow think about his way of doing it.

He solveth in mysterious ways...

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Shadow wrote:

$2^x = 50 \iff 2^{2x}=(2^x)^2=50^2=2500\quad 2^{2x-2}={1\over 4} 2^{2x} =625$

Huh, now it's better, thanks anyway. :wink:

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