Question Number 221896 by fantastic last updated on 12/Jun/25
$${If}\:\mathrm{log}\underset{\mathrm{10}} {\:}\mathrm{7}={a}\:,{then}\:\mathrm{log}\underset{\mathrm{10}} {\:}\left(\frac{\mathrm{1}}{\mathrm{70}}\right)=? \\ $$
Answered by fantastic last updated on 12/Jun/25
$$\mathrm{log}\underset{\mathrm{10}} {\:}\mathrm{7}={a} \\ $$$$\:{so}\:\mathrm{10}^{{a}} =\mathrm{7}\: \\ $$$${or}\:\mathrm{70}\:=\mathrm{10}^{{a}} .\mathrm{10} \\ $$$$=\mathrm{10}^{\left({a}+\mathrm{1}\right)} \\ $$$${So}\:\frac{\mathrm{1}}{\mathrm{70}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}^{\left({a}+\mathrm{1}\right)} } \\ $$$$=\mathrm{10}^{−\left({a}+\mathrm{1}\right)} \\ $$$${then} \\ $$$$\:\mathrm{log}\underset{\mathrm{10}} {\:}\left(\frac{\mathrm{1}}{\mathrm{70}}\right) \\ $$$$=\mathrm{log}\underset{\mathrm{10}} {\:}\mathrm{10}^{−\left({a}+\mathrm{1}\right)} \\ $$$$=−\left({a}+\mathrm{1}\right)\mathrm{log}\underset{\mathrm{10}} {\:}\mathrm{10} \\ $$$$=−\left({a}+\mathrm{1}\right)×\mathrm{1} \\ $$$$=−\left({a}+\mathrm{1}\right) \\ $$$${am}\:{I}\:{wrong}??? \\ $$
Commented by mahdipoor last updated on 12/Jun/25
$${nope}\:, \\ $$$${log}\frac{\mathrm{1}}{\mathrm{70}}=−{log}\mathrm{70}=−\left({log}\mathrm{10}+{log}\mathrm{7}\right)=−\left(\mathrm{1}+{a}\right) \\ $$
Commented by fantastic last updated on 12/Jun/25
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