# Thread: If c=log_30 3, d=log_30 5. Then express log_30 8 in terms of c and d.

1. ## If c=log_30 3, d=log_30 5. Then express log_30 8 in terms of c and d.

$\displaystyle log_{30} 3=c$

$\displaystyle log_{30} 5=d$

Then $\displaystyle log_{30} 8=$

A)$\displaystyle 2(1-c-d)$
B)$\displaystyle 3(1+c+d)$
C)$\displaystyle 3(1+c-d)$
D)$\displaystyle 3(1-c-d)$

I don't know how it could be expressed like this?

2. ## Re: If c=log_30 3, d=log_30 5. Then express log_30 8 in terms of c and d.

Use rules of logarithms. Since it is multiple choice, there are only four possible answers. Check each one until you find one that works.

\displaystyle \begin{align*}2(1-c-d) & = 2\left(1-\log_{30} 3 - \log_{30} 5\right) \\ & = 2\left(\log_{30} 30 - \log_{30} 3 - \log_{30} 5\right) \\ & = 2\left(\log_{30} \dfrac{30}{3} - \log_{30} 5\right) \\ & = 2\left(\log_{30} 10 - \log_{30} 5\right) \\ & = 2\left(\log_{30} \dfrac{10}{5}\right) \\ & = 2\log_{30} 2 = \log_{30}2^2 = \log_{30} 4 \neq \log_{30} 8\end{align*}

So, the answer is not A. You try for B, C, and D.

3. ## Re: If c=log_30 3, d=log_30 5. Then express log_30 8 in terms of c and d.

Hello, NameIsHidden!

$\displaystyle \text{Given: }\,\log_{30}3\,=\,c,\;\log_{30}5\,=\,d.\quad \text{Find }\,\log_{30}8$

. . $\displaystyle (A)\;2(1-c-d) \qquad (B)\;3(1+c+d) \qquad (C)\;3(1+c-d) \qquad (D)\;3(1-c-d)$

Since $\displaystyle \log(8) \,=\, \log(2^3) \,=\, 3\log(2)$,
I would look for a way to get a "2", using 3 and 5 (and 30)
with only multiplication and/or division.

I see that: $\displaystyle 2 \:=\:\frac{30}{3\cdot5}$

Then:-$\displaystyle \log_{30}(2) \:=\:\log_{30}{\left(\frac{30}{3\cdot 5}\right) \:=\:\log_{30}(30) - \log_{30}(3\cdot5) \:=\:1 - (\log_{30}3 + \log_{30}5)$

. . . . . . $\displaystyle =\;1 - \log_{30}3 - \log_{30}5\:=\:1 - c - d$

Therefore: .$\displaystyle \log_{30}8 \;=\;3\log_{30}2 \;=\;3(1-c-d)\;\text{ . . . Answer (D)}$

4. ## Re: If c=log_30 3, d=log_30 5. Then express log_30 8 in terms of c and d.

Originally Posted by Soroban
Hello, NameIsHidden!

Since $\displaystyle \log(8) \,=\, \log(2^3) \,=\, 3\log(2)$,
I would look for a way to get a "2", using 3 and 5 (and 30)
with only multiplication and/or division.

I see that: $\displaystyle 2 \:=\:\frac{30}{3\cdot5}$

Then:-$\displaystyle \log_{30}(2) \:=\:\log_{30}{\left(\frac{30}{3\cdot 5}\right) \:=\:\log_{30}(30) - \log_{30}(3\cdot5) \:=\:1 - (\log_{30}3 + \log_{30}5)$

. . . . . . $\displaystyle =\;1 - \log_{30}3 - \log_{30}5\:=\:1 - c - d$

Therefore: .$\displaystyle \log_{30}8 \;=\;3\log_{30}2 \;=\;3(1-c-d)\;\text{ . . . Answer (D)}$
Thank you very much.

Originally Posted by SlipEternal
Use rules of logarithms. Since it is multiple choice, there are only four possible answers. Check each one until you find one that works.

\displaystyle \begin{align*}2(1-c-d) & = 2\left(1-\log_{30} 3 - \log_{30} 5\right) \\ & = 2\left(\log_{30} 30 - \log_{30} 3 - \log_{30} 5\right) \\ & = 2\left(\log_{30} \dfrac{30}{3} - \log_{30} 5\right) \\ & = 2\left(\log_{30} 10 - \log_{30} 5\right) \\ & = 2\left(\log_{30} \dfrac{10}{5}\right) \\ & = 2\log_{30} 2 = \log_{30}2^2 = \log_{30} 4 \neq \log_{30} 8\end{align*}

So, the answer is not A. You try for B, C, and D.
But some questions consume a lot of time when doing like this checking each options. Anyway thank you.