Question

Pls solve this for me Log√35+log√2-log√7

Answer 1

Hello!  

Applying the base 10 logarithmic properties, we have:

$\log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _{10}\left(\sqrt{35} )+\log _{10}\left(\sqrt{2})=\log _{10}\left(\sqrt{35}*\sqrt{2})$

If we have:

$Log\sqrt{35}\:+\:log\sqrt{2}-log\sqrt{7}$

So:

$\log _{10}\left(\sqrt{35}\sqrt{2}\right)-\log _{10}\left(\sqrt{7}\right) =\:?$

$\log _{10}\left(\sqrt{70}\right)-\log _{10}\left(\sqrt{7}\right) =\:?$

Applying the base 10 logarithmic properties, we have:

$\log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\dfrac{a}{b}\right)$

$\log _{10}\left(\sqrt{70} \right)-\log _{10}\left(\sqrt{7} \right)=\log _{10}\left(\dfrac{\sqrt{70}}{\sqrt{7}}\right)$

$\log _{10}\left(\dfrac{\sqrt{70}}{\sqrt{7}}\right) =\:?$

$= \dfrac{\sqrt{70}}{\sqrt{7}}$

$= \dfrac{\sqrt{2}*\sqrt{5}*\sqrt{7}\!\!\!\!\!\!\!\dfrac{\hspace{0.4cm}}{~}}{\sqrt{7}\!\!\!\!\!\!\!\dfrac{\hspace{0.4cm}}{~}}$

$= \sqrt{2}*\sqrt{5}$

$= \sqrt{10}$

So:

$= \log _{10}\left(\sqrt{10}\right)$

$=\log _{10}\left(10^{\frac{1}{2}}\right)$

Applying the base 10 logarithmic properties, we have:

$\log _a\left(x^b\right)=b* \log _a\left(x\right)$

$\log _{10}\left({10}^\frac{1}{2} \right)=\dfrac{1}{2} * \log _{10}\left(10\right)$

$= \dfrac{1}{2} * 1$

$= \boxed{\frac{1}{2}}\:\:or\:\:\boxed{\:0.5}\end{array}}\qquad\checkmark$

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I Hope this helps, greetings ... DexteR! =)