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Question Number 92488 by hmamarques1994@gmail.com last updated on 07/May/20

(3x)^(log_b 3) = (5x)^(log_b 5) x = ?

$$\:\left(\mathrm{3x}\right)^{\mathrm{log}_{\mathrm{b}} \:\mathrm{3}} \:=\:\left(\mathrm{5x}\right)^{\mathrm{log}_{\mathrm{b}} \:\mathrm{5}} \\ $$$$\: \\ $$$$\:\mathrm{x}\:=\:? \\ $$

Commented by jagoll last updated on 07/May/20

(3)^(log _b (3x)) = (5)^(log _b (5x)) log _((5x)) (3x) = log _3 (5) log _3 (x){1−log _3 (5)} = (log _3 (5)−1)(log _3 (5)+1) log _3 (x) = −log _3 (15) ⇒ x = (1/(15))

$$\left(\mathrm{3}\right)^{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{3x}\right)} \:=\:\left(\mathrm{5}\right)^{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{5x}\right)} \\ $$$$\mathrm{log}\:_{\left(\mathrm{5x}\right)} \left(\mathrm{3x}\right)\:=\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right) \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\left\{\mathrm{1}−\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)\right\}\:=\:\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)−\mathrm{1}\right)\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)+\mathrm{1}\right) \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\:=\:−\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{15}\right) \\ $$$$\Rightarrow\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{15}} \\ $$

Commented by jagoll last updated on 07/May/20

cool man ������

Commented by hmamarques1994@gmail.com last updated on 07/May/20

Mito! :)

$$\left.\mathrm{Mito}!\::\right) \\ $$

Commented by john santu last updated on 07/May/20

joos ������

Commented by Ar Brandon last updated on 07/May/20

Which laws of logarithm did you apply, please ? ��

Commented by john santu last updated on 07/May/20

a^(log _b (c)) = c^(log _b (a))

$$\mathrm{a}^{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{c}\right)} \:=\:\mathrm{c}^{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{a}\right)} \: \\ $$

Commented by Ar Brandon last updated on 07/May/20

�� really ? thanks

Commented by john santu last updated on 07/May/20

log _b (3x) log _b (3)=log _b (5x)log _b (5) ((log _b (3x))/(log _b (5x))) = ((log _b (5))/(log _b (3))) ((1+log _3 (x))/(log _3 (5)+log _3 (x))) = log _3 (5) 1+log _3 (x)=(log _3 (5))^3 +log _3 (5)log _3 (x) log _3 (x){1−log _3 (5)} =(log _3 (5))^2 −1

$$\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{3x}\right)\:\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{3}\right)=\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{5x}\right)\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{5}\right) \\ $$$$\frac{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{3x}\right)}{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{5x}\right)}\:=\:\frac{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{5}\right)}{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{3}\right)} \\ $$$$\frac{\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)}\:=\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right) \\ $$$$\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)=\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)\right)^{\mathrm{3}} +\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right) \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\left\{\mathrm{1}−\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)\right\}\:=\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)\right)^{\mathrm{2}} −\mathrm{1} \\ $$

Commented by john santu last updated on 07/May/20

of course. you can proof it

$$\mathrm{of}\:\mathrm{course}.\:\mathrm{you}\:\mathrm{can}\:\mathrm{proof}\:\mathrm{it} \\ $$

Commented by Ar Brandon last updated on 07/May/20

thank you ��

Commented by jagoll last updated on 07/May/20

������

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