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$100(1-\frac{12 \cdot 0.85^{a+1}}{12 \cdot 0.85^a})$

As seen here(Or pic below), this evaluates to 15.

enter image description here

There is one step I am worried is wrong, can I just rewrite the exponentiations over like this?

$100(1-\frac{12 \cdot (a+1) \cdot \log 0.85}{12 \cdot a \cdot \log 0.85})$

Because in the end I end up with the following
$\frac{100}{a}$
Instead of 15.

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The proposed rewriting is not correct. You could rewrite, for example, $0.85^{a+1}$ as $10^{(a+1)\log(0.85)}$ if by $\log$ you mean logarithm to base $10$, or replace $10$ by $e$ if by $\log$ you mean the natural logarithm. The rewriting would not be helpful, however. You can simplify directly, $u^{a+1}/u^a=u$. –  André Nicolas Oct 10 '11 at 19:19

1 Answer 1

up vote 1 down vote accepted

No, you took the logarithm of one piece of the expression, which will not preserve anything about it (and note, even if you took the logarithm of the entire expression, you would then be looking to prove it was equal to $\log(15)$, not $15$). For example, if we had the expression $$1-\frac{10^5}{10^2}$$ (which is equal to $1-1000=-999$) it's incorrect to claim it's equal to $$1-\frac{5\log(10)}{2\log(10)}=1-\frac{5}{2}=-\frac{3}{2}$$ However, in your problem, what you can do is note that $$0.85^{a+1}=0.85\times 0.85^a$$ and divide.

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What algebraic rules is that last bit based on? Neat stuff, would love to read more about it. –  Algific Oct 10 '11 at 19:25
    
It is one of the basic properties of exponentiation. –  Zev Chonoles Oct 10 '11 at 19:26
    
That way of using this rule never occurred to me, thanks. –  Algific Oct 10 '11 at 19:30

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