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Question Number 92179 by otchereabdullai@gmail.com last updated on 05/May/20

$${\left(\frac{\mathrm{x}}{12}\right)}^{{\log }_{\sqrt{3}}\mathrm{x}}={\left(\frac{\mathrm{x}}{18}\right)}^{{\log }_{\sqrt{2}}\mathrm{x}}$$$$\mathrm{find}\mathrm{x}$$

Commented by john santu last updated on 05/May/20

$${\left(x\right)}^{\log {}_{\sqrt{3}}\left(\frac{x}{12}\right)}={\left(x\right)}^{\log {}_{\sqrt{2}}\left(\frac{x}{18}\right)}$$$$(x-1)(\log {}_{\sqrt{3}}\left(\frac{x}{12}\right)-\log {}_{\sqrt{2}}\left(\frac{x}{18}\right))=0$$$$\left(i\right)x=1$$$$\left(ii\right)2\log {}_3\left(\frac{x}{12}\right)=2\log {}_2\left(\frac{x}{18}\right)$$$$\log {}_3\left(x\right)-\log {}_3\left(12\right)=$$$$\log {}_2\left(x\right)-\log {}_2\left(18\right)$$$$\log {}_3\left(x\right)-\log {}_2\left(x\right)=\log {}_3\left(12\right)-\log {}_2\left(18\right)$$$$\log {}_3\left(x\right)\{1-\log {}_2\left(3\right)\}=$$$$\log {}_3\left(6\right)+\log {}_3\left(2\right)-\log {}_2\left(6\right)-\log {}_2\left(3\right)$$$$\log {}_3\left(x\right)\{1-\log {}_2\left(3\right)\}=2.\{1-\log {}_2\left(3\right)\}$$$$\Rightarrow x=3^2=9$$$$solutionx=1\wedge x=9$$

Commented by jagoll last updated on 05/May/20

good ☺☺☺✔️

Commented by otchereabdullai@gmail.com last updated on 05/May/20

$$\mathrm{thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}!$$

Commented by otchereabdullai@gmail.com last updated on 05/May/20

$$\mathrm{but}\mathrm{sir}\mathrm{pls}\mathrm{x}=9\mathrm{do}\mathrm{not}\mathrm{satisfy}$$

Commented by otchereabdullai@gmail.com last updated on 05/May/20

$$\mathrm{and}\mathrm{please}\mathrm{the}\mathrm{base}\mathrm{in}\mathrm{the}\mathrm{question}$$$$\mathrm{was}\sqrt{3}\mathrm{and}\sqrt{2}\mathrm{in}\left(\mathrm{ii}\right)\mathrm{but}\mathrm{pls}\mathrm{why}\mathrm{base}3\mathrm{and}$$$$2\mathrm{please}\mathrm{i}\mathrm{want}\mathrm{to}\mathrm{understand}\mathrm{thanks}$$$$\mathrm{sir}$$

Commented by otchereabdullai@gmail.com last updated on 05/May/20

$$\mathrm{am}\mathrm{much}\mathrm{greatful}\mathrm{sir}\mathrm{God}\mathrm{bless}\mathrm{you}!$$

Commented by john santu last updated on 05/May/20

$${\left(9\right)}^{\log {}_{\sqrt{3}}\left(\frac{9}{12}\right)}={\left(9\right)}^{\log {}_3{\left(\frac{3}{4}\right)}^2}$$$${\left(3\right)}^{\log {}_3{\left(\frac{3}{4}\right)}^4}=\frac{81}{256}\left(\mathrm{Lhs}\right)$$$$\mathrm{Rhs}{\left(9\right)}^{\log {}_{\sqrt{2}}\left(\frac{9}{18}\right)}={\left(9\right)}^{\log {}_2\left(\frac{1}{4}\right)}$$$$={\left(3\right)}^{\log {}_2\left(\frac{1}{16}\right)}=\frac{1}{81}$$$$\mathrm{Lhs}\ne \mathrm{Rhs}$$$$\mathrm{x}=1\mathrm{only}\mathrm{solution}$$$$$$

Commented by john santu last updated on 05/May/20

$$\mathrm{we}\mathrm{know}\mathrm{property}\mathrm{of}\mathrm{logarithm}$$$${\mathrm{a}}^{\log {}_{\mathrm{b}}\left(\mathrm{c}\right)}={\mathrm{c}}^{\log {}_{\mathrm{b}}\left(\mathrm{a}\right)}$$

Commented by john santu last updated on 05/May/20

$$\log {}_{{\mathrm{a}}^{\mathrm{n}}}\left(\mathrm{b}\right)=\frac{1}{\mathrm{n}}\log {}_{\mathrm{a}}\left(\mathrm{b}\right)$$$$\log {}_{3^{\frac{1}{2}}}\left(\mathrm{x}\right)=2\log {}_3\left(\mathrm{x}\right)$$

Commented by otchereabdullai@gmail.com last updated on 05/May/20

$$\mathrm{Am}\mathrm{much}\mathrm{greatful}\mathrm{thank}\mathrm{you}\mathrm{sair}!$$$$\mathrm{please}\mathrm{sir}\mathrm{my}\mathrm{final}\mathrm{question}\mathrm{is}\mathrm{on}\mathrm{how}$$$$\mathrm{you}\mathrm{got}\mathrm{the}(\mathrm{x}-1)$$

Commented by jagoll last updated on 05/May/20

$$\mathrm{sir}\mathrm{if}{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^{\mathrm{g}\left(\mathrm{x}\right)}=({\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^{\mathrm{h}\left(\mathrm{x}\right)}$$$$\mathrm{then}(\mathrm{f}\left(\mathrm{x}\right)-1)(\mathrm{g}\left(\mathrm{x}\right)-\mathrm{h}\left(\mathrm{x}\right))=0$$

Commented by otchereabdullai@gmail.com last updated on 05/May/20

$$\mathrm{a}\mathrm{have}\mathrm{really}\mathrm{enjoy}\mathrm{your}\mathrm{lesson}$$$$\mathrm{God}\mathrm{richly}\mathrm{bless}\mathrm{you}$$

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