Question

let log(D^2/C^3*A) when logA=3 logB=2 logD=5
a)0.549
b)4.44
c)1
d)19

Answer 1

\[\begin{align*}\log\left(\frac{D^2}{C^3A}\right)&=\log D^2-\log\left(C^3A\right)\\
&=2\log D-\left(\log C^3+\log A\right)\\
&=2\log D-3\log C-\log A\\
&=\cdots
\end{align*}\]

Answer 2

i did that but i got none of the four answers that they give you. please explain more

Answer 3

Do you know what \(\log C\) is?

Answer 4

logC is 2

Answer 5

Ah, so that was a typo..
\[\begin{align*}\log\left(\frac{D^2}{C^3A}\right)&=\log D^2-\log\left(C^3A\right)\\
&=2\log D-\left(\log C^3+\log A\right)\\
&=2\log D-3\log C-\log A\\
&=2(5)-3(2)-3\\
&=10-6-3\\
&=\cdots
\end{align*}\]

Answer 6

yeah it was. thanks for the help

Answer 7

you're welcome