Question-57607

Question Number 57607 by Gulay last updated on 08/Apr/19

Commented by Gulay last updated on 08/Apr/19

$$\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$

Commented by MJS last updated on 08/Apr/19

used rules: log_b a =((ln a)/(ln b)) log_b a^r =rlog_b a (a)^(1/n) =a^(1/n) o log_(√2) 16 =((ln 16)/(ln (√2))) ln 16 =ln 2^4 =4ln 2 ln (√2) =ln 2^(1/2) =((ln 2)/2) ((ln 16)/(ln (√2)))=((8ln 2)/(ln 2))=8 ...

$$\mathrm{used}\:\mathrm{rules}: \\ $$$$\mathrm{log}_{{b}} \:{a}\:=\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:{b}} \\ $$$$\mathrm{log}_{{b}} \:{a}^{{r}} \:={r}\mathrm{log}_{{b}} \:{a} \\ $$$$\sqrt[{{n}}]{{a}}={a}^{\frac{\mathrm{1}}{{n}}} \\ $$$${o} \\ $$$$\mathrm{log}_{\sqrt{\mathrm{2}}} \:\mathrm{16}\:=\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{ln}\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{ln}\:\mathrm{16}\:=\mathrm{ln}\:\mathrm{2}^{\mathrm{4}} \:=\mathrm{4ln}\:\mathrm{2} \\ $$$$\mathrm{ln}\:\sqrt{\mathrm{2}}\:=\mathrm{ln}\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{ln}\:\sqrt{\mathrm{2}}}=\frac{\mathrm{8ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{8} \\ $$$$… \\ $$

Commented by maxmathsup by imad last updated on 08/Apr/19

log_(√2) (16) =((ln(16))/(ln((√2)))) =((ln(2^4 ))/((1/2)ln(2))) =((8ln(2))/(ln(2))) =8 also log_(√3) (9) =((ln(9))/(ln((√3)))) =((ln(3^2 ))/((1/2)ln(3))) =((4ln(3))/(ln(3))) =4 ⇒ log_(√2) (16)−log_(√3) (9) =8−4 =4 we remember that log_a (x)=((ln(x))/(ln(a)))

$${log}_{\sqrt{\mathrm{2}}} \left(\mathrm{16}\right)\:=\frac{{ln}\left(\mathrm{16}\right)}{{ln}\left(\sqrt{\mathrm{2}}\right)}\:=\frac{{ln}\left(\mathrm{2}^{\mathrm{4}} \right)}{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)}\:=\frac{\mathrm{8}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{2}\right)}\:=\mathrm{8}\:\:{also} \\ $$$${log}_{\sqrt{\mathrm{3}}} \left(\mathrm{9}\right)\:=\frac{{ln}\left(\mathrm{9}\right)}{{ln}\left(\sqrt{\mathrm{3}}\right)}\:=\frac{{ln}\left(\mathrm{3}^{\mathrm{2}} \right)}{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)}\:=\frac{\mathrm{4}{ln}\left(\mathrm{3}\right)}{{ln}\left(\mathrm{3}\right)}\:=\mathrm{4}\:\Rightarrow \\ $$$${log}_{\sqrt{\mathrm{2}}} \left(\mathrm{16}\right)−{log}_{\sqrt{\mathrm{3}}} \left(\mathrm{9}\right)\:=\mathrm{8}−\mathrm{4}\:=\mathrm{4}\:\:\:\:{we}\:{remember}\:{that}\:{log}_{{a}} \left({x}\right)=\frac{{ln}\left({x}\right)}{{ln}\left({a}\right)} \\ $$

Answered by $@ty@m last updated on 10/Apr/19

log _(√2) 16−log _(√3) 9 =log _(√2) ((√2))^8 −log _(√3) ((√3))^4 =8−4 =4

$$\mathrm{log}\:_{\sqrt{\mathrm{2}}} \mathrm{16}−\mathrm{log}\:_{\sqrt{\mathrm{3}}} \mathrm{9} \\ $$$$=\mathrm{log}\:_{\sqrt{\mathrm{2}}} \left(\sqrt{\mathrm{2}}\right)^{\mathrm{8}} −\mathrm{log}\:_{\sqrt{\mathrm{3}}} \left(\sqrt{\mathrm{3}}\right)^{\mathrm{4}} \\ $$$$=\mathrm{8}−\mathrm{4} \\ $$$$=\mathrm{4} \\ $$

Commented by Gulay last updated on 08/Apr/19

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 08/Apr/19

Good approach!...Minor error. log _(√2) 16−log _(√3) 9 =log _(√2) ((√2))^8 −log _(√3) ((√3))^4 =8−4=4

$${Good}\:{approach}!…{Minor}\:{error}. \\ $$$$\mathrm{log}\:_{\sqrt{\mathrm{2}}} \mathrm{16}−\mathrm{log}\:_{\sqrt{\mathrm{3}}} \mathrm{9} \\ $$$$=\mathrm{log}\:_{\sqrt{\mathrm{2}}} \left(\sqrt{\mathrm{2}}\right)^{\mathrm{8}} −\mathrm{log}\:_{\sqrt{\mathrm{3}}} \left(\sqrt{\mathrm{3}}\right)^{\mathrm{4}} \\ $$$$=\mathrm{8}−\mathrm{4}=\mathrm{4} \\ $$$$ \\ $$

Commented by $@ty@m last updated on 10/Apr/19

$${Thanks}\:{for}\:{correction}. \\ $$

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