It's actually pretty simple

Algebra Level 4

If log ⁡ 4 n 40 3 = log ⁡ 3 n 45 \log _{ 4n }{ 40\sqrt { 3 } \quad =\quad \log _{ 3n }{ 45 } } , find the value of n 3 {n}^{3} .

Source: MA θ 1991 \theta \quad 1991


The answer is 75.

Shashank Rammoorthy by Shashank Rammoorthy , 20, India

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1 solution

Let log ⁡ 4 n 40 3 = log ⁡ 3 n 45 = k \log _{ 4n }{ 40\sqrt { 3 } \quad =\quad \log _{ 3n }{ 45 } \quad =\quad k }

Therefore, ( 4 n ) k = 40 3 ( 3 n ) k = 45 ( 3 4 ) k = 45 40 3 ⇒ k = 3 / 2 { (4n) }^{ k }=40\sqrt { 3 } \\ { (3n) }^{ k }=45\\ { (\frac { 3 }{ 4 } ) }^{ k }=\frac { 45 }{ 40\sqrt { 3 } } \Rightarrow k=3/2

Subsequently, n 3 {n}^{3} = 75.

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Moderator note:

Note that since you manipulated the equations ad potentially introduced new roots, you should still go back and check that ( n , k ) (n, k) is indeed a solution to the original problem.

That's an interesting question. Thanks for sharing!

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