Exponential equation with three summands

Question

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.

Hint: notice that $10^{2x} = (10^x)^2$ ... and think of quadratic equations. — Old John, Jan 05 '13 at 12:32
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 05 '13 at 13:04

score 2 · Accepted Answer · answered Jan 05 '13 at 12:33

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 $x = 2$.

score 1 · Answer 2 · answered Jan 05 '13 at 12:34

1

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

answered Jan 05 '13 at 12:34

Martin Ueding

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Exponential equation with three summands

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