$2^{x-3} + \frac {15}{2^{3-x}} = 256$

Question

$$2^{x-3} + \frac {15}{2^{3-x}} = 256$$

Find the unknown $x$.

My attempt:

We know that $x^y . x^b = x^{y+b}$.

$$2^x . 2^{-3} + 15. 2^{-3+x} = 2^8$$

and

$$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8$$

From here, we get

$$2^x + 15 = 2^8$$

However, I'm stuck at here and waiting for your kindest helps.

Thank you.

Your "from here we get $2^x+ 15= 2^8$" is wrong. You have $(2^x)(2^{-3})+ (15)(2^{-3})(2^x)= 2^8$, factoring out $(2^{-3})(2^x)$ gives $(2^{-3})(2^x)(1+ 15)= 2^{-3}(2^x)(16)= 2^8$. Since $16= 2^4$, that is $2(2^x)= 2^8$ so $2^x= 2^8/2= 2^7$. — user247327, Jan 09 '18 at 18:36

score 4 · Answer 1 · answered Jan 09 '18 at 18:30

4

writing $$\frac{2^x}{8}+\frac{15}{8}2^x=2^8$$ so $$2^x\left(\frac{1}{8}+\frac{15}{8}\right)=2^8$$ Can you finish?

answered Jan 09 '18 at 18:30

Dr. Sonnhard Graubner

95,283

score 1 · Answer 2 · edited Jan 09 '18 at 18:45

1

hint

If we put $$t=2^{x-3}, $$

the equation becomes

$$t+15t=256$$

or $$t=16$$

$$2^{x-3} = 16$$

$$x=7$$

edited Jan 09 '18 at 18:45

Fly by Night

32,272

answered Jan 09 '18 at 18:30

hamam_Abdallah

62,951

Can we solve the questions which like this with your method? It made it easier I think. – Jan 09 '18 at 18:31
3

It looks like you read the question wrong, or it was typed wrong. – Fly by Night Jan 09 '18 at 18:31
@FlybyNight It was typed correctly. – Jan 09 '18 at 18:32
1

According to how the question is written, we actually get $t+\frac {15}{\frac 1t}=256$... – abiessu Jan 09 '18 at 18:32
What if we solve this question by putting $t$? – Jan 09 '18 at 18:33
It must be $x = 7$ – Jan 09 '18 at 18:35
Post-correction is a little more than a "hint" ;-) – Fly by Night Jan 09 '18 at 18:35
@FlybyNight What do you think about this method? – Jan 09 '18 at 18:36
and can we apply this method for all? – Jan 09 '18 at 18:36
@Enzo Now that it has been corrected, it's fine. It won't work in general. But a lot of Maths is like that: some things work some of the time, other things work the rest of the time. :-/ – Fly by Night Jan 09 '18 at 18:37
@FlybyNight Do they have to be same? I mean like if it been $2^{x-3} + \frac {15}{2^{6-x}} = 256$ Could we apply this method? – Jan 09 '18 at 18:38
@Enzo For this, use the fact that $\frac{1}{2^{6-x}} = 2^{x-6} = 2^{-3}\cdot 2^{x-3}$. That gives $2^{x-3} + 15 \cdot 2^{-3} \cdot 2^{x-3} = 256$. Putting $t = 2^{x-3}$ gives $t + \frac{15}{8} t = 256$, etc. – Fly by Night Jan 09 '18 at 18:41

score 0 · Answer 3 · answered Jan 09 '18 at 18:29

0

Hint:

As $2^{-y}=\dfrac1{2^y}$

$$\dfrac1{2^{3-x}}=2^{-(3-x)}=?$$

answered Jan 09 '18 at 18:29

lab bhattacharjee

274,582

score 0 · Answer 4 · answered Jan 09 '18 at 18:39

0

$2^{x-3} + 15\cdot 2^{x-3} = 256;$

$2^{x-3}(1+15) =256;$

$2^{x-3} =256/(16)= 16;$

$x-3=4$ , by inspection.

$x=7.$

Note:

$\dfrac{1}{2^{3-x}}= 2^{x-3}$.

answered Jan 09 '18 at 18:39

Peter Szilas

20,344
2
17
28

fleablood · Answer 5 · 2018-01-09T18:48:44.993

"$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8"$

"From here, we get"

"$2^x + 15 = 2^8$"

Uh....What? No, you don't.

You get

$2^x . 2^{-3}*2^3 + 15. 2^{-3}2^3 . 2^x = 2^8*2^3"$

$2^x + 15*2^x = 2^{8+3}$

So $16 *2^x = 2^{11}$

$2^{x+4} = 2^{11}$

$x+4 = 11$

$x = 7$.

====

I'd find it easier to do. Let $y = x-3$ so

$2^y + \frac 15{2^{-y}} = 256$

$2^y + 15*2^y = 256$

$2^y(1 + 15) = 256$

$2^y(16)= 256$

$2^y = \frac {256}{16} = \frac {2^8}{2^4} = 2^4$

$y = x-3 = 4$

$x = 3$.

=====

Someone else suggested: Let $t = 2^{x-3}$ so

$t + \frac {15}{\frac 1t} = 256$ so

$t + 15t = 256$ etc....

$2^{x-3} + \frac {15}{2^{3-x}} = 256$

5 Answers5