Question

- Q1) Prove that: 2log (15÷18) - log (25÷162)+ Log(4÷9) =log 2​

Answer 1

See below for proof.

Explanation:

Given logarithmic equation:

`2\log\left((15)/(18)\right)-\log\left((25)/(162)\right)+\log\left((4)/(9)\right)=\log \left(2\right)`

We can use Log Laws to prove that the left side of the equation equals the right side of the equation.

`\textsf{Apply the Power law:} \quad n\log_ax = \log_ax^n`

`\implies \log\left((15)/(18)\right)^2-\log\left((25)/(162)\right)+\log\left((4)/(9)\right)`

`\textsf{Apply the exponent rule:} \quad \left((a)/(b)\right)^c=(a^c)/(b^c)`

`\implies \log\left((15^2)/(18^2)\right)-\log\left((25)/(162)\right)+\log\left((4)/(9)\right)`

`\implies \log\left((225)/(324)\right)-\log\left((25)/(162)\right)+\log\left((4)/(9)\right)`

`\textsf{Apply the Quotient law:} \quad \log_ax - \log_ay=\log_a \left((x)/(y)\right)`

`\implies \log\left(((225)/(324))/((25)/(162))\right)+\log\left((4)/(9)\right)`

`\implies \log\left((225)/(324)\cdot(162)/(25)\right)+\log\left((4)/(9)\right)`

`\implies \log\left((9)/(2)\right)+\log\left((4)/(9)\right)`

`\textsf{Apply the Product law:}\quad \log_ax + \log_ay=\log_axy`

`\implies \log\left((9)/(2) \cdot (4)/(9)\right)`

`\implies \log\left((4)/(2) \right)`

`\implies \log \left(2\right)`

Hence proving that the left side of the equation equals log(2).