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$$a = \frac{3^{m-1}}{3^{-m}}, b = 3^{1-m}$$

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!

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  • $\begingroup$ Well, that's in terms of $k$, not in terms of $a$ so it can't be complete. $\endgroup$ – fleablood Apr 24 '18 at 22:18
  • $\begingroup$ $3^{m-1} \ne k^{-1}$. $3^{m-1} = \frac k3$. $\endgroup$ – fleablood Apr 24 '18 at 22:40
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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, $$ a=\frac{3^{m-1}}{3^{-m}}=\frac{3^m\cdot 3^m}{3}=\frac{k^2}{3}=\frac{1}{3}\frac{9}{b^2}=\frac{3}{b^2} $$

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Notice that:

$$b = 3^{1-m} \Rightarrow b = \frac{3}{3^{m}} \Rightarrow 3^m = \frac{3}{b}.$$

Moreover:

$$a = \frac{3^{m-1}}{3^{-m}} = 3^{m-1}\cdot3^{m} = \frac{3^{m} \cdot 3^{m}}{3} = \frac{\left(\frac{3}{b}\right)^2}{3} = \frac{3}{b^2}.$$

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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.

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You've made, I think, a precedence error:

  • Correct: $3^{m-1} = 3^{(m-1)}$
  • Incorrect: $3^{m-1} = (3^m)^{-1}$

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!

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