Exponential Equations: Solving for Unknown Variables

[projection]

Homework Statement

2^{2x}-3\times2^{x+2}+32=0

The Attempt at a Solution

i have no clue as to how to solve this...if someone has a tutorial on another website of something on these type of questions...that would be helpful...can someone provide some insight as to how approach this problem?

 

[shawshank]

u can re-write as:

0 = (2^x)^2 - 3* 2^X * 2^2 + 32

 

[rocomath]

(2^x)^2-12\cdot(2^x)+32=0

How about now? Remember that, a^{x+y}=a^xa^y Make a "substitution" if necessary.

 

[projection]

thanks!

i see how it is done...i will try to do a number of these...will post if need more help.

many thanks rocophysics and shawshank

 

[projection]

(27\cdot3^x)^x=27^x\cdot3^\frac{1}{x}

when i solve it...i end up with the answers x=0 or x=-1...the answer sheet says its just x=0...i did it multiple times...keep getting the same answer...which is right?

 

[rocomath]

u=2^x

u^2-12u+32=0

(u-4)(u-8)=0

2^x=4 \ \ \ 2^x=8

x=\frac{\ln 4}{\ln 2} \ \ \ x=\frac{\ln 8}{\ln 2}

Lol, omg! New problem, sorry. I was like wtf ... I'm not even getting 0 :p

 

[rocomath]

(27\cdot3^x)^x=27^x\cdot3^\frac{1}{x}

Simplifies to ...

3^{x^2}=3^{\frac 1 x}

I'm getting 1 as my only answer.

 

[projection]

ohh crap...my bad...the answer i was looking at was for the pervious one.

here is how i did it...tell me where i went wrong:

(3^3\cdot3^x)^x=3^3x\cdot3^(-x)

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

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

x^2+3x=2x

x^2+x=0

x(x+1)=0

x=0 or x=-1

so i did something really wrong...did i violate a rule??

 

[rocomath]

27^x\cdot3^{x^2}=27^x\cdot3^{\frac 1 x}

3^{x^2}=3^{\frac 1 x}

x^2=\frac 1 x

x^3-1=0

 

[projection]

ok...so i can't make the exponent 1/x into -x??

 

[rocomath]

ok...so i can't make the exponent 1/x into -x??

You can, but why do it? Also, 27^x cancels nicely so it's just a cubic function with one real solution.

 

[VietDao29]

ok...so i can't make the exponent 1/x into -x??

You can, but why do it? Also, 27^x cancels nicely so it's just a cubic function with one real solution.

No, he cannot do it. 1/x cannot be change to -x.

You seem to be confusing between the two:

\frac{1}{a} = a ^ {-1}, this is a fine manipulation. :)

However, \left( a ^ {\frac{1}{x}} = \sqrt[x]{a} \right) \neq \left( a ^ {-x} = \frac{1}{a ^ x} \right) The two are completely different.

You are wrong at the very first step.. Now, let's re-do it using ricophysics' way. :)

 

[rocomath]

Oh! Oops, I did not mean to mislead. I accidently read an exponent of -x, rather than a change to -x. Thanks VietDao! :-]