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

1 Answer

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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(√(35) )+\log _(10)\left(√(2))=\log _(10)\left(√(35)*√(2))

If we have:


Log√(35)\:+\:log√(2)-log√(7)

So:


\log _(10)\left(√(35)√(2)\right)-\log _(10)\left(√(7)\right) =\:?


\log _(10)\left(√(70)\right)-\log _(10)\left(√(7)\right) =\:?

Applying the base 10 logarithmic properties, we have:


\log _c\left(a\right)-\log _c\left(b\right)=\log _c\left((a)/(b)\right)


\log _(10)\left(√(70) \right)-\log _(10)\left(√(7) \right)=\log _(10)\left((√(70))/(√(7))\right)


\log _(10)\left((√(70))/(√(7))\right) =\:?


= (√(70))/(√(7))


= \frac{√(2)*√(5)*√(7)\!\!\!\!\!\!\!\frac{\hspace{0.4cm}}{~}}{√(7)\!\!\!\!\!\!\!\frac{\hspace{0.4cm}}{~}}


= √(2)*√(5)


= √(10)

So:


= \log _(10)\left(√(10)\right)


=\log _(10)\left(10^{(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}^(1)/(2) \right)=(1)/(2) * \log _(10)\left(10\right)


= (1)/(2) * 1


= \boxed{(1)/(2)}\:\:or\:\:\boxed{\:0.5}\end{array}}\qquad\checkmark

___________________________

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


User Wynn
by
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