1951 AMC 12 Problem 22

Attempt Problem 22 of the 1951 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1951 AMC 12 solutions, or check the answer key.

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22.

The values of

a

in the equation

\log_{10}(a^2-15a)=2

are:

\dfrac{15\pm\sqrt{233}}{2}

20,

-5

\dfrac{15\pm\sqrt{305}}{2}

\pm20

None of these

Convert the logarithmic equation to exponential form

Solve

a^2-15a=100

The equation is equivalent to

a^2-15a=10^2=100,

\begin{aligned} &a^2-15a-100\\ &\quad=(a-20)(a+5)\\ &\quad=0. \end{aligned}

Both

a=20

and

a=-5

make the logarithm’s argument

100,

so both are valid.

Thus, the correct answer is B.

Problem 22 in other years