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

1.	If 2^a = 3^6 = 6^c then show that 1 by c = 1 by a + 1 by b
Answer» {tex}\\huge {Q. If\\ 2^a = 3^b = 6^c,\\ Show\\ that,\\ \\frac1c=\\frac{1}{\\frac{a+1}{b}}?}{/tex}\t{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}\tPlease use brackets and mathematical signs in the message of your questions.

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