2020 AIME II Problem 3

Question

The value of

x

that satisfies

\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}

can be written as

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Change of base:

\log_{2^x} 3^{20} = \frac{20 \log 3}{x \log 2},

and similarly for the right side

The equation reduces to

\frac{20}{x} = \frac{2020}{x + 3},

which is linear in

x

By the change-of-base formula,

\log_{2^x} 3^{20} = \frac{20 \log 3}{x \log 2}

and

\log_{2^{x+3}} 3^{2020} = \frac{2020 \log 3}{(x + 3) \log 2}.

Cancelling the common factor

\frac{\log 3}{\log 2}

leaves

\frac{20}{x} = \frac{2020}{x + 3}.

Cross-multiplying gives

20x + 60 = 2020x,

so

2000x = 60

and

x = \frac{3}{100}.

Thus

m + n = 3 + 100 = 103.

Problem 3 in other years