Question Number 92488 by hmamarques1994@gmail.com last updated on 07/May/20
$$\:\left(\mathrm{3x}\right)^{\mathrm{log}_{\mathrm{b}} \:\mathrm{3}} \:=\:\left(\mathrm{5x}\right)^{\mathrm{log}_{\mathrm{b}} \:\mathrm{5}} \\ $$$$\: \\ $$$$\:\mathrm{x}\:=\:? \\ $$
Commented by jagoll last updated on 07/May/20
$$\left(\mathrm{3}\right)^{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{3x}\right)} \:=\:\left(\mathrm{5}\right)^{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{5x}\right)} \\ $$$$\mathrm{log}\:_{\left(\mathrm{5x}\right)} \left(\mathrm{3x}\right)\:=\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right) \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\left\{\mathrm{1}−\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)\right\}\:=\:\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)−\mathrm{1}\right)\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)+\mathrm{1}\right) \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\:=\:−\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{15}\right) \\ $$$$\Rightarrow\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{15}} \\ $$
Commented by jagoll last updated on 07/May/20
cool man ������
Commented by hmamarques1994@gmail.com last updated on 07/May/20
$$\left.\mathrm{Mito}!\::\right) \\ $$
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
$$\mathrm{a}^{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{c}\right)} \:=\:\mathrm{c}^{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{a}\right)} \: \\ $$
Commented by Ar Brandon last updated on 07/May/20
�� really ? thanks
Commented by john santu last updated on 07/May/20
$$\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{3x}\right)\:\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{3}\right)=\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{5x}\right)\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{5}\right) \\ $$$$\frac{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{3x}\right)}{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{5x}\right)}\:=\:\frac{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{5}\right)}{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{3}\right)} \\ $$$$\frac{\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)}\:=\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right) \\ $$$$\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)=\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)\right)^{\mathrm{3}} +\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right) \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\left\{\mathrm{1}−\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)\right\}\:=\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)\right)^{\mathrm{2}} −\mathrm{1} \\ $$
Commented by john santu last updated on 07/May/20
$$\mathrm{of}\:\mathrm{course}.\:\mathrm{you}\:\mathrm{can}\:\mathrm{proof}\:\mathrm{it} \\ $$
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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