# Sloving exponential equations

1. Feb 21, 2008

### projection

1. The problem statement, all variables and given/known data

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

3. 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?

2. Feb 21, 2008

### shawshank

u can re-write as:

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

3. Feb 21, 2008

### 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.

4. Feb 21, 2008

### 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

5. Feb 21, 2008

### 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?

6. Feb 21, 2008

### 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

7. Feb 21, 2008

### 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.

Last edited: Feb 21, 2008
8. Feb 21, 2008

### 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??

9. Feb 21, 2008

### rocomath

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

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

$$x^2=\frac 1 x$$

$$x^3-1=0$$

10. Feb 21, 2008

### projection

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

11. Feb 21, 2008

### rocomath

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

12. Feb 22, 2008

### VietDao29

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. :)

13. Feb 22, 2008

### rocomath

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