Question

simply log10( 145/8)-log10(3/2)+log10(54/29)​

Answer 1

$\textbf{To simplify:}$

$\mathsf{log_{10}\left(\dfrac{145}{8}\right)-log_{10}\left(\dfrac{3}{2}\right)+log_{10}\left(\dfrac{54}{29}\right)}$

$\textbf{Solution:}$

$\boxed{\mathsf{Quotient\;rule:\;\;\;log_a\dfrac{M}{N}=log_aM-log_aN}}$

$\boxed{\mathsf{Product\;rule:\;\;\;log_aM\,N=log_aM+log_aN}}$

$\textsf{Consider,}$

$\mathsf{log_{10}\left(\dfrac{145}{8}\right)-log_{10}\left(\dfrac{3}{2}\right)+log_{10}\left(\dfrac{54}{29}\right)}$

$\textsf{Using quotient rule, we get}$

$\mathsf{=log_{10}\left(\dfrac{\dfrac{145}{8}}{\dfrac{3}{2}}\right)+log_{10}\left(\dfrac{54}{29}\right)}$

$\mathsf{=log_{10}\left(\dfrac{145}{8}{\times}\dfrac{2}{3}}\right)+log_{10}\left(\dfrac{54}{29}\right)}$

$\textsf{Using product rule,}$

$\mathsf{=log_{10}\left(\dfrac{145}{8}{\times}\dfrac{2}{3}{\times}\dfrac{54}{29}\right)}$

$\mathsf{=log_{10}\left(\dfrac{5}{4}{\times}\dfrac{1}{3}{\times}\dfrac{54}{1}\right)}$

$\mathsf{=log_{10}\left(\dfrac{5}{4}{\times}18\right)}$

$\mathsf{=log_{10}\left(\dfrac{5}{2}{\times}9\right)}$

$\mathsf{=log_{10}\left(\dfrac{45}{2}\right)}$

$\textbf{Answer:}$

$\mathsf{log_{10}\left(\dfrac{45}{2}\right)}$

$\textbf{Find more:}$

x^2 + y^2= 25xy, then prove that 2 log(x+y)=3log3+logx+logy

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If 2log(x-y)=logx +logy, prove that 2log(x+y)= log5+logx +logy

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Answer 2

Answer:

sir ne Jo answer Diya vo theek Hain

maine check kr liya.

hope it's helps