Question

rewrite as a single logarithm.
2+log_4(a)+1/2log_4(b)-3log_4(c)-log_4(d)

what do i do with the 2

Answer 1

You could try writing 2 as a log base 4.

Answer 2

Still a bit confused.
I know the final answer... it's log_4(16a(b^1/2)/c^3d

Answer 3

\(\log_4 x = 2\)

What is x?

Answer 4

ah so x=2^4
so x = 16?

Answer 5

Correct.
Now rewrite \(2\) as \(\log_4 16\)
Then you will have an expression with all logs base 4.
Then you can use the rules of logs to write a single logarithm.

Answer 6

\(2+\log_4 a+\dfrac{1}{2}\log_4 b - 3 \log_4 c - \log_4 d\)

\(=\log_4 16+\log_4 a+\dfrac{1}{2}\log_4 b - 3 \log_4 c - \log_4 d\)

Answer 7

i see thank you :)

Answer 8

You're welcome.
Do you know how to continue?

Answer 9

yes i do, thank you very much :) was confused with the "2"

Answer 10

\(=\log_4 16+\log_4 a+\log_4 b^{\frac{1}{2}} - (\log_4 c^3 + \log_4 d)\)

\(=\log_4 16ab^{\frac{1}{2}} - \log_4 c^3d\)

\(=\log_4 \dfrac{16ab^{\frac{1}{2}}}{c^3d}\)

BTW, the answer you know is correct is indeed correct.

Answer 11

thank you!