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May 24th, 2017, 11:50 AM   #1
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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.
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May 24th, 2017, 12:23 PM   #2
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Quote:
Originally Posted by Opposite View Post
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?
Try this according to your logic
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Last edited by dthiaw; May 24th, 2017 at 12:25 PM.
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May 24th, 2017, 01:53 PM   #3
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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 $\log_{10}(3) \approx 0.4771$
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May 24th, 2017, 04:22 PM   #4
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Try this according to your logic
BIG OOpppps disregard that 4 factor
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May 25th, 2017, 09:32 AM   #5
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@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 $\log_{10}(2) \approx 0.3010$ ?
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May 25th, 2017, 09:48 AM   #6
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Originally Posted by Opposite View Post
@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 $\log_{10}(2) \approx 0.3010$ ?
The values of $\log_{10}(2)$ and $\log_{10}(3)$ come from a calculator or a log table ... were the values given to you in the problem statement?

How was $\log_6(216) = 3$ involved with the base 10 log question?
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May 25th, 2017, 10:42 AM   #7
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Quote:
Originally Posted by skeeter View Post
The values of $\log_{10}(2)$ and $\log_{10}(3)$ come from a calculator or a log table ... were the values given to you in the problem statement?

How was $\log_6(216) = 3$ involved with the base 10 log question?
I know nothing about Log, that's how it's involved.

The test only gave the first two. But I was only allowed a basic computer calculator. How would I calculate this on my own?
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May 25th, 2017, 11:13 AM   #8
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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?
Without getting too detailed, here are the main log properties one should have learned in Algebra 2 or Precalculus ...

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 ...
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May 25th, 2017, 12:19 PM   #9
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Thankyou. Is there a website you recommend with a lesson on this?
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May 25th, 2017, 12:26 PM   #10
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Thankyou. Is there a website you recommend with a lesson on this?
do a google search on properties of logarithms ... you'll have to choose a site or video that best fits your style of learning
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