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I had a simple looking math problem the other day:

Solve for $y(x) = 0$: $$ 10^{2x} - 101 \cdot 10^x + 100 = 0$$

Since I have three summands, I cannot just put them to either side of the equation and apply $\log_{10}$ to it. And I cannot see how I could factor this to get it into a product to use $\log_{10}$.

Mathematica gave me $x = 2$ as a result which seems correct. How can I find the solution to this one? It is supposed to be an elementary problem for an applied math class.

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Have you considered substituting something? –  barto Jan 5 '13 at 12:29
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Hint: notice that $10^{2x} = (10^x)^2$ ... and think of quadratic equations. –  Old John Jan 5 '13 at 12:32
    
How could I not see this? Thanks! –  queueoverflow Jan 5 '13 at 12:35
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Something else to look out for in this sort of question: sometimes they might give an equation that starts $10^{2x+1}\dots$ and then you have to notice that this can be written as $10\times (10^x)^2 \dots$ –  Old John Jan 5 '13 at 13:04
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2 Answers

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The key here is to note that $10^{2x} = (10^x)^2$. So we have $$10^{2x} - 101\cdot 10^x + 100 = (10^x)^2 - 101\cdot 10^x + 100.$$ Now letting $y = 10^x$, we are trying to solve $$y^2 - 101y + 100 = 0.$$ This has solutions $y = 1$ and $y = 100$ which correspond to $x = 0$ and $x = 2$.

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It is a quadratic equation in $10^x$. So that would make it: $$ 10^x = \frac{101}{2} \pm \sqrt{\frac{101^2}{4} - 100} $$ $$ \implies 10^x = 1 \lor 10^x = 100 \iff x = 0 \lor x = 2 $$

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