Question

Express the following logarithm as the sum of logarithms with no products, powers, or quotients.

Answer 1

`f(x) = 2\log_4x + 3\log_4(x + 9) - 8\log_4(x - 1)`

Explanation:

In the following answer I'll use the following logarithm laws:

`\log_m(a * b) = \log_ma + \log_mb\\\\\log_m(\frac ab) = \log_ma - \log_mb\\\\\log_m(a^b) = b * \log_ma`

Now for the answer:

`f(x) = \log_4\left((x^2(x + 9)^3)/((x - 1)^8)\right)\\\\\to f(x) = \log_4\left(x^2(x + 9)^3\right) - \log_4\left((x - 1)^8\right)\\\\\to f(x) = \log_4\left(x^2\right) + \log_4\left((x + 9)^3\right) - \log_4\left((x - 1)^8\right)\\\\\to f(x) = 2\log_4x + 3\log_4(x + 9) - 8\log_4(x - 1)`

Answer 2

`\begin{aligned}\\&f(x)=\log_4\left((x^2(x+9)^3)/((x-1)^8)\right)\\&f(x)=\log_4(x^2(x+9)^3)-\log_4{(x-1)^8\\&f(x)=\log_4x^2+\log_4(x+9)^3-8\log_4{(x-1)\\&f(x)=2\log_4x+3\log_4(x+9)-8\log_4{(x-1)\end{aligned}`

second option