Question

Question #0bfd7

Answer 1

`1/2log(36)+2log(3)+1=log(540)` (assuming `log` means `log_10`)

Explanation:

First, we can use the following identity:
`alog_x(b)=log_x(b^a)`

This gives:
`1/2log(36)+2log(3)+1=log(36^(1/2))+log(3^2)+1=`

`=log(6)+log(9)+1`

Now we can use the multiplication identity:
`log_x(a)+log_x(b)=log_x(a*b)`

`log(6)+log(9)+1=log(6*9)+1=log(54)+1`

I'm unsure if this is what the question is asking for, but we can also bring the `1` into the logaritm. Assuming that `log` means `log_10`, we can rewrite the `1` like so:
`log(54)+1=log(54)+log(10)`

Now we can use the same multiplication identity as before to get:
`=log(54*10)=log(540)`