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May 24, 2019

Real numbers problem - C.M.I - U.G - 2019

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] .Find all real numbers x for which
$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$  

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C.M.I (Chennai mathematical institute ) U.G-2019

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Real Numbers

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]8 out of 10

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Excursion in Mathematics by Bhaskaracharya Pratisthana[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

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It is of the the form of  $\frac{a^3+b^3}{a^2b+b^2a}$ . Do you observe ?  where a=\(2^x\) b=\(3^x\)

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\(\frac{a^3+b^3+a^2b+b^2a}{a^2b+b^2a}\)=\(\frac{13}{6}\) \(\Rightarrow\frac{a^2+b^2}{ab}=\frac{13}{6}\) \(\frac{a}{b}+\frac{b}{a}=\frac{13}{6}\) let x=a/b then it is a quadratic equation    

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\(6x^2-13x+6=0\) on solving we get x=3/2 or 2/3 now replace x by a/b

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so again  putting value of a and b we get final ans is +1 and -1 

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

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Similar Problem

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