Triple Log

Algebra Level 3

If 9 log ⁡ a = 4 log ⁡ ( a + b ) = 6 log ⁡ b ^9\log a= {}^4\log(a+b)={}^6\log b what is a b \frac{a}{b}

1 2 ( 3 − 1 ) \frac{1}{2}(\sqrt{3}-1) 1 5 ( 2 − 1 ) \frac{1}{5}(\sqrt{2}-1) 1 2 ( 5 + 1 ) \frac{1}{2}(\sqrt{5}+1) 1 5 ( 2 + 1 ) \frac{1}{5}(\sqrt{2}+1) 1 6 ( 6 + 1 ) \frac{1}{6}(\sqrt{6}+1) 1 2 ( 5 − 1 ) \frac{1}{2}(\sqrt{5}-1)

Achmad Damanhuri by Achmad Damanhuri , 26, Indonesia

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1 solution

Let x = log ⁡ 9 a = log ⁡ 4 ( a + b ) = log ⁡ 6 b x=\log_9 a=\log_4 (a+b)=\log_6 b , then a = 9 x , b = 6 x , a + b = 4 x a=9^x, b=6^x,a+b=4^x the equation becomes: 9 x + 6 x = 4 x 9^x+6^x=4^x Observe that 4 , 6 , 9 4,6,9 is a geometric sequence, and our goal is to find a b = ( 3 2 ) x \dfrac{a}{b}=\left(\dfrac{3}{2}\right)^x , divide both sides by 4 4 : ( 9 4 ) x + ( 3 2 ) x = 1 \left(\dfrac{9}{4}\right)^x+\left(\dfrac{3}{2}\right)^x=1 Let u = ( 3 2 ) x ( u > 0 ) u=\left(\dfrac{3}{2}\right)^x (u>0) u 2 + u − 1 = 0 u = 5 − 1 2 u^2+u-1=0 \\ u=\dfrac{\sqrt{5}-1}{2} Therefore, a b = u = 5 − 1 2 \dfrac{a}{b}=u=\dfrac{\sqrt{5}-1}{2}

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