1984 AHSME Problems/Problem 14

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Problem

The product of all real roots of the equation $x^{\log_{10}{x}}=10$ is

$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }-1 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }10^{-1} \qquad \mathrm{(E) \ } \text{None of these}$

Solution

Take the logarithm base $x$ of both sides to get $\log_{10}x=\log_x10$. Use the change of base formula to get $\log_{10}x=\frac{\log_{10}10}{\log_{10}x}=\frac{1}{\log_{10}x}$. Therefore, $(\log_{10}x)^2=1$ and $\log_{10}x=\pm1$. From $\log_{10}x=1$, we have $x=10^1=10$ and from $\log_{10}x=-1$ we have $x=10^{-1}=\frac{1}{10}$. The product is thus $10\times\frac{1}{10}=1, \boxed{\text{A}}$.

See Also

1984 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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