Помогите,пожалуйста, вычислить. Подробно.

Question 1

Question 2

2)→ 4/3 ; 4)→ 0

Question 3

2)
(Loq₇ 14 -(1/3)*Log₇ 56 ) / (Log₆ 30 -(1/2)* Log₆ 150 ) =
( Loq₇ 2+Loq₇ 7 -(1/3)*(Log₇ 7 + Log₇ 2³ ) )/
(Log₆ 6+Log₆ 5 -(1/2)*( Log₆ 6 + Log₆ 5² ) ) =
( Loq₇ 2+1 -1/3 - Log₇ 2 ) / (1+ Log₆ 5 -1/2 - Log₆ 5 ) = (2/3) /(1/2) = 4/3 .
=======
4)
(3Log₇ 2 -(1/2)*Log₇ 64) / (4Log₅ 2 +(1/3)*Log₅ 27) =
(3Log₇ 2 -(1/2)*Log₇ 2⁶) / (4Log₅ 2 +(1/3)*Log₅ 3³) =
(3Log₇ 2 -(1/2)*6Log₇ 2 ) / (4Log₅ 2 +(1/3)*3Log₅ 3) =0 /5Log₅ 2 =0.

Question 4

Решите задачу:

$\dfrac{\log_714- \frac{1}{3}\log_756 }{\log_630-0.5\log_6150} = \dfrac{\log_714-\log_7 \sqrt[3]{56} }{\log_630-\log_6 \sqrt{150} } = \dfrac{\log_7 \frac{14}{ \sqrt[3]{56} } }{\log_6 \frac{30}{ \sqrt{150} } } =\\ \\ \\ = \dfrac{\log_7 \frac{14}{2 \sqrt[3]{7} } }{\log_6 \frac{30}{5 \sqrt{6} } } = \dfrac{\log_7(7^{1- \frac{1}{3} })}{\log_6(6^{1- \frac{1}{2} })} = \dfrac{1- \frac{1}{3} }{1- \frac{1}{2} } = \dfrac{6-2}{6-3} = \dfrac{4}{3}$

$\dfrac{3\log_72-0.5\log_764}{4\log_52+ \frac{1}{3}\log_527 } = \dfrac{\log_72^3-\log_764^{ \frac{1}{2} }}{\log_52^4+\log_5 \sqrt[3]{27} } = \dfrac{\log_78-\log_78}{\log_516+\log_53} =0$

oganesbagoyan_zn · Answer 1 · 2018-05-10T00:17:56+0000

2)
(Loq₇ 14 -(1/3)*Log₇ 56 ) / (Log₆ 30 -(1/2)* Log₆ 150 ) =
( Loq₇ 2+Loq₇ 7 -(1/3)*(Log₇ 7 + Log₇ 2³ ) )/
(Log₆ 6+Log₆ 5 -(1/2)*( Log₆ 6 + Log₆ 5² ) ) =
( Loq₇ 2+1 -1/3 - Log₇ 2 ) / (1+ Log₆ 5 -1/2 - Log₆ 5 ) = (2/3) /(1/2) = 4/3 .
=======
4)
(3Log₇ 2 -(1/2)*Log₇ 64) / (4Log₅ 2 +(1/3)*Log₅ 27) =
(3Log₇ 2 -(1/2)*Log₇ 2⁶) / (4Log₅ 2 +(1/3)*Log₅ 3³) =
(3Log₇ 2 -(1/2)*6Log₇ 2 ) / (4Log₅ 2 +(1/3)*3Log₅ 3) =0 /5Log₅ 2 =0.