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MISCELLANEOUS EXERCISE 6 (II) · Q183

Q.If log⁡2a4=log⁡2b6=log⁡2c3k\dfrac{\log_2 a}{4}=\dfrac{\log_2 b}{6}=\dfrac{\log_2 c}{3k} and a3b2c=1a^3b^2c=1, find the value of kk.

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Let log⁡2a4=log⁡2b6=log⁡2c3k=m\dfrac{\log_2a}{4}=\dfrac{\log_2b}{6}=\dfrac{\log_2c}{3k}=m.

Then log⁡2a=4m, log⁡2b=6m, log⁡2c=3km\log_2a=4m,\ \log_2b=6m,\ \log_2c=3km, i.e. a=24m, b=26m, c=23kma=2^{4m},\ b=2^{6m},\ c=2^{3km}.

Given a3b2c=1a^3b^2c=1: 212m⋅212m⋅23km=224m+3km=20=12^{12m}\cdot2^{12m}\cdot2^{3km}=2^{24m+3km}=2^0=1 (since a3b2c=1=20a^3b^2c=1=2^0). …

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