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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

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

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

$${log}_{\sqrt{2}}\left(16\right)=\frac{ln\left(16\right)}{ln\left(\sqrt{2}\right)}=\frac{ln\left(2^4\right)}{\frac{1}{2}ln\left(2\right)}=\frac{8ln\left(2\right)}{ln\left(2\right)}=8also$$$${log}_{\sqrt{3}}\left(9\right)=\frac{ln\left(9\right)}{ln\left(\sqrt{3}\right)}=\frac{ln\left(3^2\right)}{\frac{1}{2}ln\left(3\right)}=\frac{4ln\left(3\right)}{ln\left(3\right)}=4\Rightarrow$$$${log}_{\sqrt{2}}\left(16\right)-{log}_{\sqrt{3}}\left(9\right)=8-4=4werememberthat{log}_a\left(x\right)=\frac{ln\left(x\right)}{ln\left(a\right)}$$

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

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

Commented by Gulay last updated on 08/Apr/19

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

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

$$Goodapproach!...Minorerror.$$$$\log {}_{\sqrt{2}}16-\log {}_{\sqrt{3}}9$$$$=\log {}_{\sqrt{2}}{\left(\sqrt{2}\right)}^8-\log {}_{\sqrt{3}}{\left(\sqrt{3}\right)}^4$$$$=8-4=4$$$$$$

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

$$Thanksforcorrection.$$

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