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If 2^a = 3^6 = 6^c …

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If 2^a = 3^6 = 6^c then show that 1 by c = 1 by a + 1 by b

    • Posted by Wildheart Devil 4 years, 5 months ago

    CBSE > Class 09 > Mathematics
    • 1 answers

    Rajan Kumar Pasi 4 years, 5 months ago

    {tex}\huge {Q. If\ 2^a = 3^b = 6^c,\ Show\ that,\ \frac1c=\frac{1}{\frac{a+1}{b}}?}{/tex}

    {tex}\huge{Sol:\ Let's\ first\ take\ ,\ 2^a = 3^b }\\ \large{\ \text{(Applying log on both the sides)}}\\ \huge{=>\ \log\ 2^a=\log\ 3^b}\\ \huge{=>\ a\log2=b\log3}\\ \huge{=>\ a={b{\log3\over\log2}}}\\ {/tex} {tex}\huge{Similarly\ ,\ 3^b=6^c }\\ \large{\ \text{(Applying log on both the sides)}}\\ \huge{=>\ \log\ 3^b=\log\ 6^c}\\ \huge{=>\ b\log3=c\log6}\\ \huge{=>\ c={b{\log3\over\log6}}}\\ {/tex}
    {tex}\huge{LHS\ ,\ \frac1c = \frac{\log6}{b\log3} }\\ \huge{RHS\ ,\ \frac1a+\frac1b= \frac{\log2}{b\log3}+\frac1b }\\ \huge{=>\frac1b\Bigg( \frac{\log2}{\log3}+1\Bigg) }\\ \huge{=>\frac1b\Bigg( \frac{\log2+\log3}{\log3}\Bigg) }\\ \huge{=>\frac1b\Bigg( \frac{\log(2\times 3)}{\log3}\Bigg) }\\ \huge{=>\frac1b\Bigg( \frac{\log6}{\log3}\Bigg) }\\ {/tex} {tex}\huge{Thus,\ LHS=RHS}\\ \huge{\frac1c =\ \frac1a+\frac1b= \huge{\frac1b\Bigg( \frac{\log6}{\log3}\Bigg) }\\ }\\ {/tex}

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