1. ## neg. exponent?

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?

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

3. 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.

4. 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 $z=2^x$

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

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

7. 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:
$y=-3,-4$.

8. 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.

9. 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 $x = \frac{log 3}{log 2}$

10. 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.

11. 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.