3x-log-b-3-5x-log-b-5-x-

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

{(3 x)}^{\log_{b} 3} = {(5 x)}^{\log_{b} 5}

x = ?

Commented by jagoll last updated on 07/May/20

{(3)}^{\log_{b} (3 x)} = {(5)}^{\log_{b} (5 x)}

\log_{(5 x)} (3 x) = \log_{3} (5)

\log_{3} (x) {1 - \log_{3} (5)} = (\log_{3} (5) - 1) (\log_{3} (5) + 1)

\log_{3} (x) = - \log_{3} (15)

\Rightarrow x = \frac{1}{15}

Commented by jagoll last updated on 07/May/20

cool man ��

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

Mito! :)

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)}

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

�� really ? thanks

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

\log_{b} (3 x) \log_{b} (3) = \log_{b} (5 x) \log_{b} (5)

\frac{\log_{b} (3 x)}{\log_{b} (5 x)} = \frac{\log_{b} (5)}{\log_{b} (3)}

\frac{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

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

of course . you can proof 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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