1956 AMC 12 Problem 20

Attempt Problem 20 of the 1956 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 1956 AMC 12 solutions, or check the answer key.

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20.

(0.2)^x=2

and

\log 2=0.3010,

then the value of

x

to the nearest tenth is:

-10.0

-0.5

-0.4

-0.2

10.0

Take common logarithms of both sides

Use

\log(0.2)=\log2-1=-0.6990

Taking common logarithms gives

x\log(0.2)=\log2.

Since

\log(0.2)=\log2-1=-0.6990,

x=\frac{0.3010}{-0.6990}\approx-0.4306.

To the nearest tenth, this is

-0.4.

Thus, the correct answer is C.

Problem 20 in other years