Question

Simplify:
log, (3x+4) – 7 log, r° +6log, x​

Answer 1

Step-by-step explanation:

In this case, we have the following expression:

`log(3x+4) - 7 log(r^(0)) +6log(x)`

Remember that any number whose exponent is 0 equals 1, so:

`r^(0)=1`

Then, we can write our expression as:

`log(3x+4) - 7 log(1) +6log(x)`

We know that:

`log(1)=0`

So our expression becomes:

`log(3x+4) +6log(x) \\ \\ \\ By \ property: \\ \\ nlog(m)=log(m^n) \\ \\ \\ Then: \\ \\ log(3x+4) +log(x^6) \\ \\ \\ By \ property: \\ \\ log(mn)=log(m)+log(n) \\ \\ \\ So: \\ \\ log(3x+4) +log(x^6)=log[(3x+4)(x^6)]=\boxed{log(3x^7+4x^6)}`