Spain - Math Olympiad Problem | Be Careful!

You should know this approach. Many goes WRONG! Solution

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[Limited_Light]

By the definition of logarithms (in this case, using base 2 / 3), x = log_[2 / 3](3^2 / 2^3) = ln(3^2 / 2^3) / ln (2 / 3), by Change of Base in the last.

x = (log 3^2 - log 2^3) / (log 2 - log 3) = (2 log 3 - 3 log 2) / (log 2 - log 3).

[DmytroMiller]

In my humble opinion, the solution is unnecessary overcomplicated. You simply apply logarithm to the original equation to get immediately (x+3)ln2=(x+2)ln3, which is a simple linear equation to solve.

[martinmolander]

Alternately use 2^(x+3)=3(x+2) => 2 2^(x+2)=3^(x+2) => (3/2)^(x+2)=2 => (x+2) ln (3/2)=ln 2 => x = ln 2/ ln (3/2) - 2 = 2( ln (sqrt(2))/ln(3/2) - 1)

Since ln(sqrt(2)) is approx 0, 42 - h.o.t., and ln 3/2 is approx 0, 5 - h.o.t, so the answer is about - 4*0, 08 .

[KUDIYARASAN-]

After point, it is mathematical convention to denote, zero point two nine and not twenty nine.... please fix it Sir.