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(√(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`

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