2001 SMT/Algebra Problems/Problem 7

Problem

If $\log_A B + \log_B A = 3$ and $A<B$ , find $\log_B A$ .

Solution

Let $x=\log_B A$ ; then $\frac{1}{x}+x=3$ . From here we find that $x^2-3x+1=0$ , so $x=\frac{3\pm\sqrt{5}}{2}$ . Technically both solutions are possible even if $A<B$ (because of values less than 1); however, we assume that we are considering $1<A<B$ ; then $x=\log_B A<1$ , so $x=\boxed{\frac{3-\sqrt{5}}{2}}$ .

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2001 SMT/Algebra Problems/Problem 7

Problem

Solution