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Algebra Level 5

$2\log_x a + \log_{ax} a + 3\log_b a = 0$

For $a>0$ and $b=a^2 x$ , solve the equation above.

If the solutions can be expressed as $a^c$ and $a^d$ , and $c + d$ can be expressed as $-\dfrac ef$ for coprime positive integers $e$ and $f$ , evaluate $e-f$ .

This is part of the set My Problems and THRILLER

The answer is 5.

1 solution

Chew-Seong Cheong
Oct 19, 2015

$\begin{aligned} 2\log_x a + \log_ax a + \log_{\color{#3D99F6}{b}} a & = 0 \quad \quad \small \color{#3D99F6}{b = a^2x} \\ \frac{2\log_a a}{\log_a x} + \frac{\log_a a}{\log_a (ax)} + \frac{3\log_a a}{\log_a (a^2x)} & = 0 \\ \frac{2}{\color{#3D99F6}{\log_a x}} + \frac{1}{1 + \color{#3D99F6}{\log_a x}} + \frac{3}{2 + \color{#3D99F6}{\log_a x}} & = 0 \quad \quad \small \color{#3D99F6}{\text{Let } y = \log_a x} \\ 2(1+y)(2+y) + y(2+y) + 3y(1+y) & = 0 \\ 2y^2+6y+4 + y^2+2y + 3y^2+3y & = 0 \\ 6y^2+11y+4 & = 0 \\ (2y+1)(3y+4) & = 0 \\ \Rightarrow \log_a x & = \begin{cases} - \frac{1}{2} \\ - \frac{4}{3} \end{cases} \quad \Rightarrow x = \begin{cases} a^{- \frac{1}{2}} \\ a^{- \frac{4}{3}} \end{cases} \end{aligned}$

$\Rightarrow c + d = -\frac{1}{2} - \frac{4}{3} = - \frac{11}{6} \quad \Rightarrow e-f = = 11-6 = \boxed{5}$

Exactly same way sir! Upvoted

Department 8 - 5 years, 7 months ago

same approach!

Prince Loomba - 5 years, 1 month ago

Same way sir

abhishekrocks sahoo - 5 years, 1 month ago

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The answer is 5.

1 solution

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