Can someone please explain how to solve for m and n.

[frank plaza]

So the question is 2^m = 2^n + 56. And I need to find the value of m and n. I already know the answer, which is m = 6 and n = 3. Just want to know how to solve it myself.

[frank plaza]

Also is this a rule or what

how is 2^m-n turning into 2^m / 2^n

[ancient basin]

,tex 2^{m-n} = 2^m \cdot 2^{-n}

But

x^{-n} = \frac{1}{x^n}

Hence,

2^{m-n} = 2^m \cdot 2^{-n} = 2 \cdot \frac{1}{2^n} = \frac{2^m}{2^n}

[past moth]

Gkratsia
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[frank plaza]

I see thanks. So x^-n = 1/x^n always

[ancient basin]

That's right!

[frank plaza]

Why exactly?

🤔

[ancient basin]

By definition, I would say... It's the negative exponent rules.

Although it is not a proof, you can check that for example:

$2^2\cdot 2^{-1} = 2^{2-1} = 2$

also

$2^2 \cdot 2^{-1} = 2^2 \cdot \frac{1}{2^1} = \frac{4}{2} = 2$

[past moth]

Gkratsia

[frank plaza]

I see. Thank you so much !

Makes a lot more sense now

[ancient basin]

Coming to your initial question:

$2^m = 2^n + 56 \Rightarrow 2^m - 2^n = 56$ for this to hold, m must be greater than n.

[past moth]

Gkratsia

[ancient basin]

You are welcome!

A first and naive approach would be the following

$2^m - 2^n = 64 - 8 = 2^6 - 2^3$, hence $m=6$ and $m=3$

[past moth]

Gkratsia

[unique yoke]

Take log base 2 of everything

[frank plaza]

Damn how did you assume the values of 2^m and 2^n?

[ancient basin]

But I think there is a better solution like boice said

[frank plaza]

I did, but then I got lost later on.

m = log2(2^n + 56)

something like this

[ancient basin]

$56 = 64 - 8 = 2^6 - 2^3$ I tried to find a combination of powers of two that would give me 56.

[past moth]

Gkratsia

[frank plaza]

Ohh shit that's smart

thanks though I've kind of solved it but I don't know how to write with latex

[unique yoke]

Is this problem underdefined?

[ancient basin]

But not always the correct way...

[frank plaza]

2^m = 2^n + 56
2^n(2^{m-n} - 1) = 8 x 7
2^n(2^{m-n} - 1) = 2^3(2^3-1)
m = 6, n = 3

[frank plaza]

what do you mean by underdefined?

[frank plaza]

this was my approach @ancient basin

[unique yoke]

Is there only one solution

Or is there more than one

[ancient basin]

It seems there is only one, but it's not always the case.

I like your approach better than mine

Use the dollar sign to make it latex

[frank plaza]

I don't know. This was the main question. We need to find values of m and n

[ancient basin]

$2^m = 2^n + 56$

$2^n(2^{m-n} - 1) = 8 \times 7$

$2^n(2^{m-n} - 1) = 2^3(2^3-1)$

$m = 6, n = 3$

[frank plaza]

I see, thanks!

[past moth]

Gkratsia

[unique yoke]

Yeah 1 equation two unknowns so if m and n could be non integers there are infinite solutions

[frank plaza]

damn alright, thanks though.

thanks for all the help

i'm closing this thread

.close