How would I condense this into one logarithm

Looks like you don't know the Sum and Difference of Logs!

Step-by-step explanation:

Okay, before we solve this, we need to review what is said above.

For example, how do we condense this?

Imagine: Log(2)(4)+Log(2)(8) and Log(2)(4)-Log(2)(8)- The first () is the base, and the second () is the other one. ( doesn't provide a Log symbol)

For this to work, the two logs must have same base. If not, it doesn't work.

Sum:

First: See the other () - (The second number with the parenthesis)?

Second: To condense this with the sum of logarithms, the two "Other's" are Multiplied, so that it would be Log(2)(4)+Log(2)(8)=Log(2)(32), because 4*8=32

Difference

First: See the other () - (The second number with the parenthesis)?

Second: To condense this with the sum of logarithms, the two "Other's" are Divided, so that it would be Log(2)(4)-Log(2)(8)=Log(2)(1/2), because 4/8=1/2

Did you get it? Well, its time to solve!

Before we use our recently learned Sum and Difference of Logs, one must condense the Coefficients outside of the Logs to do so.

Doing the First Term and Third Term:

4Log(2)(x+6)=Log(2)((x+6)^4)        (NOTE: The ^4 stands for "to the power of four, and (x+6)^4 is the complete "Other".)

8Log(2)(3x)=Log(2)((3x)^8)

Badda Bing Badda Boom! We're ready to use it!

Starting off:

Log(2)((x+6)^4) + Log(2)(x+2) - Log(2)((3x)^8)

= Log(2)() - Log(2)()

= Log(2)()        <<--------------- Answer

Hope this helps! If you want to simplify it, its up to you.

If you want more review, check out some other online resources. They explain it much better than I ever can.

70.30

step-by-step explanation: