May 24th, 2017, 11:50 AM  #1 
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9  Logarithms...
I took an AccuPlacer and just barely missed it. I noticed this question and when I tried researching it, something didn't match up and I'm hoping you know what I'm missing. My basic understanding is that $\displaystyle log_{6} 216 = 3$ but the question I'm trying to remember didn't seem to match... The answer is an estimate from my memory. Hoping someone could point me in a direction that might be more helpful. I'm suppose to figure out the last problem based on the answers to the first two. To state it again, the .3000 and the .4700 are based on memory, I wasn't allowed to bring notes with me. $\displaystyle log_{10} 2 = .3000$ , $\displaystyle log_{10} 3 = .4700$, $\displaystyle log_{10} 24 = ?????$ Any guess? Last edited by Opposite; May 24th, 2017 at 11:53 AM. 
May 24th, 2017, 12:23 PM  #2  
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8  Quote:
Last edited by dthiaw; May 24th, 2017 at 12:25 PM.  
May 24th, 2017, 01:53 PM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,883 Thanks: 1503 
Using the values you recall ... $\log(24)=\log( 8 \cdot 3) = \log(8 )+\log(3)=\log(2^3)+\log(3)=3\log(2)+\log(3)= 3(.3000)+0.4700 = 1.3700$ In actuality, $\log_{10}(2) \approx 0.3010$ and $\log_{10}(3) \approx 0.4771$ 
May 24th, 2017, 04:22 PM  #4 
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8  
May 25th, 2017, 09:32 AM  #5 
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9 
@Skeeter, yes this was the set of equations. $\log_{10}(2) \approx 0.3010$ and $\log_{10}(3) \approx 0.4771$ But I still don't get it. I only know the original formula, $\log_{6} 216 = 3$ and this tells me that 6*6*6 = 216.... How does $\log_{10}(2) \approx 0.3010$ ? 
May 25th, 2017, 09:48 AM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 2,883 Thanks: 1503  Quote:
How was $\log_6(216) = 3$ involved with the base 10 log question?  
May 25th, 2017, 10:42 AM  #7  
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9  Quote:
The test only gave the first two. But I was only allowed a basic computer calculator. How would I calculate this on my own?  
May 25th, 2017, 11:13 AM  #8  
Math Team Joined: Jul 2011 From: Texas Posts: 2,883 Thanks: 1503  Quote:
relationship between a logarithm and its corresponding exponential function ... $\log_b(y) = x \implies y = b^x$, $b > 0$, $b\ne 1$, and $y>0$ Properties of logs ... 1. $\log_b(1) = 0$ 2. $\log_b(x \cdot y) = \log_b(x) + \log_b(y)$ 3. $\log_b \left(\dfrac{x}{y} \right) = \log_b(x)  \log_b(y)$ 4. $\log_b(x^n) = n \cdot \log_b(x)$ If the question was to determine the value of $\log_{10}(24)$ using the given decimal values for $\log_{10}(2)$ and $\log_{10}(3)$, the problem is testing one's knowledge of the properties of logs, specifically properties #2 and #4 above ... $\log_{10}(24) = \log_{10}(8 \cdot 3) = \log_{10}(2^3 \cdot 3) = \log_{10}(2^3) + \log_{10}(3) = 3\color{red}{\log_{10}(2)} + \color{blue}{\log_{10}(3)}$ from here, one would substitute in the given decimal values for $\log_{10}(2)$ and $\log_{10}(3)$ ... $3 \times \color{red}{0.3010} + \color{blue}{0.4771}$ now use the basic calculator to evaluate ...  
May 25th, 2017, 12:19 PM  #9 
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9 
Thankyou. Is there a website you recommend with a lesson on this?

May 25th, 2017, 12:26 PM  #10 
Math Team Joined: Jul 2011 From: Texas Posts: 2,883 Thanks: 1503  

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