Multivariable Log problem

[bktheking]

Having some trouble figuring out how to go about this problem.

Let log_b 2 = a and log_b 4 = c. Find log_b 8b⁴

It wants the answer in terms of the variables a, b, and a number e.g 3a + 4 or 3(a + b).

I tried expanding log_b 8b⁴ using several properties of logs but fail to obtain an answer every time.
I just wanted to see if someone could give me some insight on how to go about this. Thanks in advance.

 

[soroban]

Hello, bktheking!

You were on your way . . . Why did you stop?

\(\displaystyle \text{Let }\log_b(2) = a\,\text{ and }\,\log_b(4) = c.\;\;\text{ Find }\log_b(8b^4)\)

\(\displaystyle \log_b(8b^4) \;=\;\log_b(8) + \log_b(b^4)\)

. . . . . . . .\(\displaystyle =\;\log_b(2\cdot4) + 4\underbrace{\log_b(b)}_{\text{This is 1}} \)

. . . . . . . .\(\displaystyle =\;\log_b(2) + \log_b(4) + 4\)

. . . . . . . .\(\displaystyle =\;a + c + 4\)

 

[pka]

Having some trouble figuring out how to go about this problem.
Let log_b 2 = a and log_b 4 = c. Find log_b 8b⁴

You know that \(\displaystyle \log_b(8)=a+b\). WHY?
Moreover, \(\displaystyle \log_b(b)=1\)

 

[bktheking]

You were on your way . . . Why did you stop?

Thanks for you quick reply! I stopped because I forgot log_b b was equal to 1.

You know that

. WHY?

Because log_b(xy) = log_bx + log_by. I also think you meant log_b (8) = a+c, correct?

One last thing I wanted to be sure of:
If you look at my original post, it said the answer had to be in terms of the variables a and b (not c). I did the problem again and got the answer 3a + 4,
because a+c = log_b 2³ > 3log_b2, correct?