$$a = \frac{3^{m-1}}{3^{-m}}, b = 3^{1-m}$$
What is the value of $a$ in terms of $b$?
I want to solve this problem using ''Letter''.
Let $3^m = k$
$$ b = 3^{1-m} \implies \frac{1}{b} = 3^{m-1} \implies \frac{1}{b} = k^{-1} $$
However, I'm not sure whetheror not it is correct.
Regards!
On
Notice that:
$$b = 3^{1-m} \Rightarrow b = \frac{3}{3^{m}} \Rightarrow 3^m = \frac{3}{b}.$$
Moreover:
$$a = \frac{3^{m-1}}{3^{-m}} = 3^{m-1}\cdot3^{m} = \frac{3^{m} \cdot 3^{m}}{3} = \frac{\left(\frac{3}{b}\right)^2}{3} = \frac{3}{b^2}.$$
On
You've made, I think, a precedence error:
If you had $(3^m)^{-1}$, you could indeed make a substitution to rewrite it as $k^{-1}$. But, alas, that's not what you actually have.
To use your approach, you need to use the exponent laws to isolate $3^m$; for example,
$$ \frac{1}{b} = 3^{m-1} = 3^m \cdot 3^{-1} = k \cdot \frac{1}{3} = \frac{k}{3} $$
Your idea of using $k$ is a good one: it allows you to simplify the two equations which may help in solving the given problem. You just have to do the algebra correctly!
On
Well, if we set $w = \frac ab$ then $a = wb$.
And $w = \frac {\frac{3^{m-1}}{3^{-m}}}{3^{1-m}} = 3^{(m-1)-(-m) - (1-m)} = 3^{3m-2}$ so $a = 3^{3m-2}b$.
Perhaps the first thing we should have done was simplify $a = \frac{3^{m-1}}{3^{-m}}= 3^{(m-1)-(-m)}= 3^{2m-1}$ and so $b = 3^{1-m}$ then $a = 3^{2m-1} = 3^{1-m}*3^{(2m-1) - (1-m)} = b*3^{3m -2}$.
but to do it your way with letters.
$k = 3^m$ so $a= \frac{3^{m-1}}{3^{-m}} = \frac {\frac k3}{\frac 1k} = \frac {k^2}3$.
And $b = 3^{1-m} = \frac 3k$
$a = \frac{3^{m-1}}{3^{-m}}=\frac {\frac k3}{\frac 1k} = \frac {k^2}3= \frac kb=\frac {3^m}b$.
Now it looks like these two answers are in conflict.
But they aren't actually. As $b= 3^{1-m}$ then $a= 3^{3m-2}*b = 3^{3m-2}3^{1-m} = 3^{2m-1}$. And $a = \frac {3^m}{b} =\frac {3^m}{3^{1-m}} = 3^{2m - 1}$.
So those two answers are in agreement.
The second answer is the one we would have gotten if we had set $v = ab$ then $a = \frac vb$. Then as $v = {\frac{3^{m-1}}{3^{-m}}}{b^{1-m}}= 3^{(m-1)-(-m) + (1-m)} = 3^{m}$ and $a = \frac {3^m}{b}$.
$a$ and $b$ are constants this shouldn't be surprising. One can always express one constant in an infinitude of ways in relation to the other.
Your relation $\frac{1}{b}=k^{-1}$ would imply $b=k$, which is false.
If $k=3^m$, then $$ b=3^{1-m}=\frac{3}{3^m}=\frac{3}{k} $$ so $k=3/b$.
On the other hand, $$ a=\frac{3^{m-1}}{3^{-m}}=\frac{3^m\cdot 3^m}{3}=\frac{k^2}{3}=\frac{1}{3}\frac{9}{b^2}=\frac{3}{b^2} $$