math question dealing with exponents

[jobberd]

Im having trouble answering this question. Could someone show me the steps to solving it?
2^2x + 2^3x = 80, solve for x
i know youre supposed to factor, but when i do that, i get
2^2x(1 + 2^x) = 80
and im lost as to what to do from there....can anyone help? (btw, i know what x is, and im sure many of you could figure it out just by guessing, but i need to know HOW to do it..) Thanks a lot

 

[Zugzwang152]

use substitution, let u = 2^x...so it becomes:

u^2 + u^3 = 80

u^3 + u^2 - 80 = 0


get it now? solve for u, then let that equal 2^x...

 

[ArmenK]

I haven't thought of a good proof yet but here is a crappy one:
<bad proof>
we know 80 = 2^4 * 5 = 2^2x * (1 + 2^x)
so we have 5 = 2^(2x-4) + 2^(3x-4) = 2^2 + 1 = 2^2 + 2^0
so you get 2x-4 = 2 or 0 => x= 2 or 3
and 3x-4 = 2 or 0 => x=2 or 4/3

x=2 is the solution for both so it is the right answer
</bad proof>

 

[Gibson486]

doesn't this involve logarithms? in other wods, you have to take the natural log (ln) of both sides.

ln(2^2x+2^3x)=ln(80)

distribute ln (this is where I am not sure, are you allowd to distrubute ln? Or can you not take ln of that term)

ln (2^2x)+ ln (2^3x)=ln 80

ln brings exponents down, so
2x ln 2 + 3x ln 2 =ln 80

Undistrubute ln 2
ln 2 (2x+3x)=ln 80
divide ln2 on both sides

5x=ln 80/ln2
solve with calculator.

 

[ArmenK]

you cannot distribute logs:

(based 10) log10 + log 10 = 2 != log (10 + 10)= 1.301

 

[Soccer55]

How about this:

We have: 2^(2x) + 2^(3x) = 80

As ArmenK pointed out, 80 = 5*(2^4), so now we have: 2^(2x) + 2^(3x) = 5*(2^4)

Take the ln of both sides to get: ln[2^(2x) + 2^(3x)] = ln[5*(2^4)]

Factor a 2^(2x) out on the left side to get: ln[2^(2x) *{1+ 2^x}] = ln[5*(2^4)]

We know that ln(a*b) = ln(a) + ln(b), so we apply that to this equation to get: ln[2^2x] + ln[1+2^x] = ln[5] + ln[2^4]

So now either ln[2^2x] = ln[5] and ln[1+2^x] = ln[2^4] OR ln[2^2x] = ln[2^4] and ln[1+2^x] = ln[5]

So we have:

2^(2x) = 5 and 1+2^x = 2^4

OR

2^2x = 2^4 and 1+2^x = 5

Then just solve for x and test your x with the original equation to find out which x works (only one should)

-Tom