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How can I solve the following equation:

$$2^{x}+\log_{10}x-2=0$$

Any help welcome. Thanks!

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Is that term in the middle supposed to be the base-$10$ logarithm of $x$? If so, can you think of some really really simple number that makes the equation work? and then think about how the functions are increasing, and why that means there can't be more than one solution? –  Gerry Myerson Feb 15 '13 at 5:15
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3 Answers 3

up vote 4 down vote accepted

Define $f(x) = 2^x + \log x - 2$ and note $f$ is increasing on the interval $(0,\infty)$ (since both $2^x$ and $\log x$) are increasing. Then there can be at most one root - it's easy to see that $f(1) = 0,$ so $x=1$ is the unique solution.

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Word for word, exactly what I would have written. Naturally, therefore, I think this is an excellent answer. –  Lubin Feb 20 '13 at 18:14
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Try a few values and see what you get for this part:

$2^{x}+$log$_{10}^{x}-2$

For example, for $x=1$, you'll get 0 as $2^x + $log$_{10}^{x} = 2+0 = 2 $

Now, if x > 1, then the $2^{x}+$log$_{10}^{x}$ will be bigger than 2 so those can be removed. For $0<x<1$, look at how the log now is negative and thus the exponent on the first term has to be greater than one. At least that how I'd look at it.

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$$2^{x}+\log_{10}x-2=0$$

$$2^{x}-2=-log_{10}x$$ Either $2^x -2\ge 0$ and$ -log_{10}x\ge0$

$2^x\ge 2 $ and $ log_{10}x \le 0$ $x\ge1 and x \lg 1$

$ x=1$

OR $2^x -2\le 0$ and$ -log_{10}x\le0$

$2^x\le 2 $ and $ log_{10}x \ge 0$

$x\le1 $and $x \ge 1$

$ x=1$

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