neg. exponent?

**Barthayn** · December 15th 2009, 02:14 PM

Hi, I have an neg. exponent in my equation below. This is what the question asks for:

7. Solve each equation. Check for extraneous roots.

$2^x + 12(2)^{-x} = 7$

How would I solve this?

**abender** · December 15th 2009, 02:16 PM

$a^{-b} = \frac{1}{a^b}$

**Barthayn** · December 15th 2009, 02:20 PM

that doesn't work for $2^x + 12(2)^{-x} = 7$ . You need to get an $a^2$ in it to solve this equation. My teacher was telling a person in my class yesterday to mulitply it by another a value, $2^x$ in this case, to cancel out the neg. exponent.

**Plato** · December 15th 2009, 02:27 PM

Originally Posted by Barthayn

7. Solve each equation. Check for extraneous roots.
$2^x + 12(2)^{-x} = 7$
How would I solve this?

$2^x + 12(2)^{-x} = 7$ is the same as $2^{2x} -7\cdot 2^x+ 12=0$ .

Can you solve $z^2-7z+12=0?$

The let $z=2^x$

**Barthayn** · December 15th 2009, 02:31 PM

can you explain to my how did you canceled out the neg. exponent?

**pickslides** · December 15th 2009, 02:37 PM

multiply the equation through by $2^x$ like this $2^x(2^x + 12(2)^{-x} = 7) \Rightarrow 2^{2x} + 12 = 7(2^x)$

**abender** · December 15th 2009, 04:43 PM

Sure it works.

$ 2^x + 12(2)^{-x} = 7 \implies 2^x + 12(\frac{1}{2^x}) = 7 \implies 2^x + \frac{12}{2^x} = 7 $ .

Seeing it this way should scream multiply both sides by the common denominator, which is $2^x$ . So,

$2^x (2^x)+ \frac{12}{2^x}(2^x) = 7 (2^x) \implies 2^{2x} + 12 = 7(2^x)$

Now, make life easier by using the assignment $y=2^x$ .

The equation then becomes $y^2 + 12 = 7y$ .

Rewrite the equation as $y^2 - 7y + 12 = 0$ .

Factoring, we get:
$y=-3,-4$ .

**abender** · December 15th 2009, 05:00 PM

Don't forget to finish the problem.

Plugging $2^x$ back in for $y$ , we get:

$ 2^x = -3 \implies \log_2{2^x} = \log_2{(-3)} \implies x = \log_2{(-3)} $
$ 2^x = -4 \implies \log_2{2^x} = \log_2{(-4)} \implies x = \log_2{(-4)} $

Clearly, both of these solutions are extraneous roots.

**Barthayn** · December 15th 2009, 05:50 PM

Originally Posted by abender

Don't forget to finish the problem.

Plugging $2^x$ back in for $y$ , we get:

$ 2^x = -3 \implies \log_2{2^x} = \log_2{(-3)} \implies x = \log_2{(-3)} $
$ 2^x = -4 \implies \log_2{2^x} = \log_2{(-4)} \implies x = \log_2{(-4)} $

Clearly, both of these solutions are extraneous roots.

No, you are incorrect. According to my answer I got help and the answer book the answers are $x = 2$ and $x = \frac{log 3}{log 2}$ . Never in my career of high school I was taught that one can put a fraction into a polynomial to make it a polynomial.

These were the steps that one has to take to get the correct answer

Let a = $2^x$

$2^x + 12(2)^{-x} - 7 = 0$
$2a + 12-a - 7 x a = 0$
$2a^2 - 7a + 12 = 0$
$(a-4)$ $(a-3)$
$a = 4$ or $a = 3$
$2^x = 4$ or $2^x = 3$
$x = 2$ or $x = \frac{log 3}{log 2}$

**mr fantastic** · December 15th 2009, 05:54 PM

Originally Posted by abender

[snip]
Factoring, we get: Mr F says: (y - 3)(y - 4).
$y=-3,-4$ . Mr F says: Therefore y = 3, 4.

[snip]

..

Originally Posted by abender

Don't forget to finish the problem.

Plugging $2^x$ back in for $y$ , we get:

$ 2^x = {\color{red}3} \implies \log_2{2^x} = \log_2{(-3)} \implies x = \log_2{({\color{red}3})} $
$ 2^x = {\color{red}4} \implies \log_2{2^x} = \log_2{(-4)} \implies x = \log_2{({\color{red}4})} $

[snip]

Minor corrections in red (you can't take the log of a negative number and get something real!!). Both solutions should now be tested in the original equation.

**mr fantastic** · December 15th 2009, 06:12 PM

The method (if not some of the details) used by abender is correct and is completely consistent with the solution given by the 'answer book'. Note:

$\log_2 4 = 2$ .

$\log_2 3$ can be written in the form yur answer uses by using the change of base rule. Alternatively, $2^x = 3 \Rightarrow \log 2^x = \log 3 \Rightarrow x \log 2 = \log 3 \Rightarrow x = \frac{\log 3}{\log 2}$ .

I will also add that the question as originally posted made no reference to a particular method, so the hint given in post #2 is completely justified.

This thread is now closed.

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