AMC 12B 2024 (Problem 8)What value of xxx satisfies (log2x)(log3x)log2x+log3x=2?\frac{(\log_2 x)(\log_3 x)}{\log_2 x+\log_3 x}=2?log2x+log3x(log2x)(log3x)=2?(A) 25\text{(A)}\;25(A)25(B) 32\text{(B)}\;32(B)32(C) 36\text{(C)}\;36(C)36(D) 42\text{(D)}\;42(D)42(E) 48\text{(E)}\;48(E)48Related TopicsCoreToolkit 48 — Logarithms: Definition and PropertiesHints (5)Hint 1Use the change of base formula. logab=1logba.\log_a b=\frac1{\log_b a}.logab=logba1.Hint 2(log2x)(log3x)log2x+log3x=2\frac{(\log_2 x)(\log_3 x)}{\log_2 x+\log_3 x}=2log2x+log3x(log2x)(log3x)=2⟹1logx2⋅1logx31logx2+1logx3=2.\Longrightarrow \frac{\frac1{\log_x2}\cdot\frac1{\log_x3}}{\frac1{\log_x2}+\frac1{\log_x3}}=2.⟹logx21+logx31logx21⋅logx31=2.Hint 3Simplify the left-hand side. 1logx2+logx3=2\frac1{\log_x2+\log_x3}=2logx2+logx31=2⟹logx2+logx3=12.\Longrightarrow \log_x2+\log_x3=\frac12.⟹logx2+logx3=21.Hint 4logx6=12.\log_x6=\frac12.logx6=21.Hint 5x1/2=6x^{1/2}=6x1/2=6⟹x=36.\Longrightarrow x=36.⟹x=36.Final Answer(C) 363636