Question

prove that log (15÷7)_log (25÷4)+log (35÷12)=0​

Answer 1

by using log x/y= log x-logy and logo.y=logo+log y formulas

Answer 2

$\log(\dfrac{15}{7})-\log(\dfrac{25}{4})+\log (\dfrac{35}{12})=0$

Step-by-step explanation:

To prove: $\log(\dfrac{15}{7})-\log(\dfrac{25}{4})+\log (\dfrac{35}{12})=0$

Properties of logarithm are

Quotient property : $\log(\dfrac{a}{b})=\log a-\log b$

Product property : $\log(ab)=\log a+\log b$

Taking LHS,

$LHS=\log(\dfrac{15}{7})-\log(\dfrac{25}{4})+\log (\dfrac{35}{12})$

Using Quotient property we get

$LHS=\log(15)-\log(7)-(\log(25)-\log(4))+\log (35)-\log(12)$

$LHS=\log(3\times 5)-\log(7)-\log(5\times 5)+\log(4)+\log (5\times 7)-\log(3\times 4)$

Using Product property we get

$LHS=\log(3)+\log(5)-\log(7)-(\log(5)+\log(5))+\log(4)+\log (5)+\log(7)-(\log(3)+\log(4))$

$LHS=\log(3)+\log(5)-\log(7)-\log(5)-\log(5)+\log(4)+\log (5)+\log(7)-\log(3)-\log(4)$

$LHS=0$

$LHS=RHS$

Hence proved.

#Learn more

Log(225/32)_log(25/81)+log(64/729)

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