1

If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. Your answer should no longer include any logarithms.

I noted that $\log_5 10=\frac{1}{\log_{10} 5}.$

I also noted that $\log_{5} 10=\log_5 2+\log_5 5=\log_5 2+1.$ I don't know how to continue, how do I finish this problem using my strategy?

Frost Bite
  • 541

5 Answers5

4

$$\log_83=P \iff \log_{2^3}3=P \iff \frac{1}{3}\log_23=P \iff \log_23=3P \iff \log_32=\frac{1}{3P}.$$

$$\log_35=Q.$$ Dividing last equalities at the end of both lines, we get: $$\log_52=\frac{1}{3PQ}$$ Also, we know $$\log_{10}5=\frac{1}{\log_510}=\frac{1}{\log_52\cdot5}=\frac{1}{\log_52+\log_55}=\frac{1}{\log_52+1}=\frac{1}{\frac{1}{3PQ}+1}=\frac{3PQ}{3PQ+1}$$

VIVID
  • 11,604
  • How did you get $\log_83=P \iff \log_32=\frac{1}{3P}?$ – Frost Bite Apr 07 '20 at 18:24
  • @FrostBite edited accordingly. Ask if still not clear. – VIVID Apr 07 '20 at 18:27
  • By meaning Dividing last equalities at the end of both lines, do you mean $\log_32=\frac{1}{3P}$ and $\log_35=Q?$ – Frost Bite Apr 07 '20 at 18:30
  • @FrostBite Exactly – VIVID Apr 07 '20 at 18:30
  • Ok, I think I get the question and way to do it a lot better now. I will review your solution thoroughly. It seems very complete. – Frost Bite Apr 07 '20 at 18:31
  • Well, the last line IMO needs to be elaborated: $10^x=5 \Rightarrow 5^{\frac{1}{x}}=10$ $\Rightarrow 5^{\frac{1}{x}-1} = 2$ $\Rightarrow$ $\log_5 2 =1/x-1$ where $x=\log_{10} 5$ – Mike Apr 07 '20 at 18:34
  • 1
    @Mike, IMO, I feel like it doesn't need to be elaborated too much. I understand how he got there because it was some of the stuff I posted that I knew about the question already. Thank you for clarifying, though. :) – Frost Bite Apr 07 '20 at 18:36
  • 1
    @Mike Well, edited accordingly :) – VIVID Apr 07 '20 at 18:38
3

\begin{align*} \log_8 3 = P\quad\implies \quad 3&=8^P=2^{3P}\\ \log_3 5 = Q\quad\implies \quad 5&=3^Q=(2^{3P})^Q=2^{3PQ} \end{align*} From the last equation it follows that $$\log_5 2=\frac1{\log_2 5}=\frac1{3PQ}.$$

ᴇɴᴅᴇʀᴍᴀɴ
  • 143
Ángel Mario Gallegos
  • 19,882
1

$$ \log_{10}5 = \frac{\log_35}{\log_310} = \frac{\log_35}{\log_35 + \log_32} = \frac{3\log_35}{3\log_35 + \log_38} = \frac{3Q}{3Q + 1/P} = \frac{3PQ}{3PQ + 1}. $$

Calum Gilhooley
  • 12,174
0

$$ \log_{10} 5 = \log_{8} 5 \times \log_{10}{8}$$

$$\log_{10}{8}=\frac{1}{\log_{8}{10}} = \frac{1}{\log_{8} 2+\log_8 5}$$

$$\log_8 5 = \log_8 3\times \log_3 5 = PQ, \log_{8} 2=\frac{1}{3}$$

$$ \log_{10} 5 = PQ\times \frac{1}{\frac{1}{3}+PQ}$$

SMA.D
  • 1,467
  • 12
  • 27
0

An attempt. But I'm not sure whether the assumption I make at the beginning is correct, or whether it is arbitrary.

We have :

$P = \log_8 3 \iff 8^P = 3$

and

$Q = log_3 5 \iff 3^Q = 5$.

Suppose that : $log_{10} 5 = kPQ$

$log_{10} 5 = kPQ$

$\rightarrow 10^{log_{10} 5} = 10^{kPQ}$

$\rightarrow 5 = 10^{kPQ}$

$\rightarrow 3^Q = 10^{kPQ}$

$\rightarrow {(8^P)}^Q = 10^{kPQ}$

$\rightarrow 8^{PQ} = 10^{kPQ}$

$\rightarrow 8^{PQ} = {(8^{\log_{8}{10}})}^{kPQ}$

$\rightarrow 8^{PQ} = {8^{(\log_{8}{10\times kPQ)}}}$

$\rightarrow PQ = \log_{8}{10\times kPQ}$

$\rightarrow 1 = \log_{8}{10\times k}$ ( Dividing by $PQ$ on both sides).

$\rightarrow k = \frac {1} { \log_{8}{10}}$

Therefore, $\log_{10}5 = k\times PQ = \frac {1} { \log_{8}{10}} \times PQ$

Symbolab gives $\frac {1} { \log_{8}{10}} \times PQ = \frac {\ln 5} {\ln 10}$

But $\frac {\ln 5} {\ln 10} = \log_{10} 5$ using the change of base formula , with base $e$.

So, number $k$ seems to do the job.