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.
$\displaystyle 2^x + 12(2)^{-x} = 7$
How would I solve this?
that doesn't work for $\displaystyle 2^x + 12(2)^{-x} = 7$. You need to get an $\displaystyle a^2$ in it to solve this equation. My teacher was telling a person in my class yesterday to mulitply it by another a value, $\displaystyle 2^x$ in this case, to cancel out the neg. exponent.
Sure it works.
$\displaystyle
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 $\displaystyle 2^x $. So,
$\displaystyle 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 $\displaystyle y=2^x$.
The equation then becomes $\displaystyle y^2 + 12 = 7y $.
Rewrite the equation as $\displaystyle y^2 - 7y + 12 = 0 $.
Factoring, we get:
$\displaystyle y=-3,-4 $.
Don't forget to finish the problem.
Plugging $\displaystyle 2^x $ back in for $\displaystyle y $, we get:
$\displaystyle
2^x = -3 \implies \log_2{2^x} = \log_2{(-3)} \implies x = \log_2{(-3)}
$
$\displaystyle
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 $\displaystyle x = 2$ and $\displaystyle x = \frac{log 3}{log 2}$. 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 = $\displaystyle 2^x$
$\displaystyle 2^x + 12(2)^{-x} - 7 = 0$
$\displaystyle 2a + 12-a - 7 x a = 0$
$\displaystyle 2a^2 - 7a + 12 = 0$
$\displaystyle (a-4)$ $\displaystyle (a-3)$
$\displaystyle a = 4$ or $\displaystyle a = 3 $
$\displaystyle 2^x = 4$ or $\displaystyle 2^x = 3$
$\displaystyle x = 2$ or $\displaystyle x = \frac{log 3}{log 2}$
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:
$\displaystyle \log_2 4 = 2$.
$\displaystyle \log_2 3$ can be written in the form yur answer uses by using the change of base rule. Alternatively, $\displaystyle 2^x = 3 \Rightarrow \log 2^x = \log 3 \Rightarrow x \log 2 = \log 3 \Rightarrow x = \frac{\log 3}{\log 2}$.
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.