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

((x/(12)))^(log_(√3) x) =((x/(18)))^(log_(√2) x) find x

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

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

(x)^(log _(√3) ((x/(12)))) = (x)^(log _(√2) ((x/(18)))) (x−1)(log _(√3) ((x/(12)))−log _(√2) ((x/(18))))=0 (i) x = 1 (ii) 2log _3 ((x/(12))) = 2log _2 ((x/(18))) log _3 (x)−log _3 (12) = log _2 (x)−log _2 (18) log _3 (x)−log _2 (x)= log _3 (12)−log _2 (18) log _3 (x){1−log _2 (3)} = log _3 (6)+log _3 (2)−log _2 (6)−log _2 (3) log _3 (x){1−log _2 (3)} = 2. {1−log _2 (3)} ⇒ x = 3^2 = 9 solution x = 1 ∧ x = 9

$$\left({x}\right)^{\mathrm{log}\:_{\sqrt{\mathrm{3}}} \:\left(\frac{{x}}{\mathrm{12}}\right)} \:=\:\left({x}\right)^{\mathrm{log}\:_{\sqrt{\mathrm{2}}} \left(\frac{{x}}{\mathrm{18}}\right)} \\ $$$$\left({x}−\mathrm{1}\right)\left(\mathrm{log}\:_{\sqrt{\mathrm{3}}} \left(\frac{{x}}{\mathrm{12}}\right)−\mathrm{log}\:_{\sqrt{\mathrm{2}}} \left(\frac{{x}}{\mathrm{18}}\right)\right)=\mathrm{0} \\ $$$$\left({i}\right)\:{x}\:=\:\mathrm{1} \\ $$$$\left({ii}\right)\:\mathrm{2log}\:_{\mathrm{3}} \left(\frac{{x}}{\mathrm{12}}\right)\:=\:\mathrm{2log}\:_{\mathrm{2}} \left(\frac{{x}}{\mathrm{18}}\right) \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left({x}\right)−\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{12}\right)\:=\: \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)−\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{18}\right)\: \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left({x}\right)−\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{12}\right)−\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{18}\right) \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left({x}\right)\left\{\mathrm{1}−\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}\right)\right\}\:= \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{6}\right)+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)−\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{6}\right)−\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}\right) \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left({x}\right)\left\{\mathrm{1}−\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}\right)\right\}\:=\:\mathrm{2}.\:\left\{\mathrm{1}−\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}\right)\right\} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{3}^{\mathrm{2}} \:=\:\mathrm{9}\: \\ $$$${solution}\:{x}\:=\:\mathrm{1}\:\wedge\:{x}\:=\:\mathrm{9} \\ $$

Commented by jagoll last updated on 05/May/20

good ☺☺☺✔️

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

thanks for your time sir!

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

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

but sir pls x= 9 do not satisfy

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

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

and please the base in the question was (√3) and (√(2 )) in (ii) but pls why base 3 and 2 please i want to understand thanks sir

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

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

am much greatful sir God bless you!

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

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

(9)^(log _(√3) ((9/(12)))) = (9)^(log _3 ((3/4))^2 ) (3)^(log _3 ((3/4))^4 ) = ((81)/(256)) (Lhs) Rhs (9)^(log _(√2) ((9/(18)))) = (9)^(log _2 ((1/4))) =(3)^(log _2 ((1/(16)))) = (1/(81)) Lhs ≠ Rhs x = 1 only solution

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

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

we know property of logarithm a^(log _b (c)) = c^(log _b (a))

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

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

log _a^n (b) = (1/n)log _a (b) log _3^(1/2) (x) = 2 log _3 (x)

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

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

Am much greatful thank you sair! please sir my final question is on how you got the (x−1)

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

Commented by jagoll last updated on 05/May/20

sir if (f(x))^(g(x)) = ((f(x))^(h(x)) then (f(x)−1)(g(x)−h(x))=0

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

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

a have really enjoy your lesson God richly bless you

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

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