My Math Forum A couple Logarithm Questions

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 December 15th, 2016, 02:48 AM #1 Newbie   Joined: Dec 2016 From: Canada Posts: 9 Thanks: 0 A couple Logarithm Questions Having a slight issue completely understanding logarithms and their relation to exponents. Log 32 ------ Log 6 I understand that you can turn the 32 into 6^2 as a common "answer"? I guess is the term because log of 32 is whatever you raise the base to the power of in exponential. So now you have. 2 Log 6 --------- Log 6 Now the logs can cancel (because I'm assuming because the base and the "answer 6" are the same, we assume the exponents are the same so they can cross off?) and I'm left with 2 as my final answer. So why exactly can I drop the two? I know it's according to that power rule but I'm confused as to why the exponent 2 now becomes the left over answer. And what is the exponential equivalent for writing 2 Log 6 -------- Log 6 I think I'm overthinking this. And one other log question I can't figure out Given LogB (3x)=25, if LogB (27)=k-3LogB (X) Sorry for the weird questions.
 December 15th, 2016, 05:30 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983 $32 \ne 6^2$ ... maybe you meant to post $\dfrac{\log{36}}{\log{6}}$ ? $\log_b(3x)=25 \implies \log_b(3) + \log_b(x) = 25 \implies \log_b(3)=25-\log_b(x)$ Multiply every term in the last equation by 3 ... $3\log_b(3) = 75 - 3\log_b(x)$ compare to your last equation $\log_b(27)=k-3\log_b(x)$ ... what is $k$? Thanks from zekea
December 15th, 2016, 06:01 AM   #3
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 Originally Posted by skeeter $32 \ne 6^2$ ... maybe you meant to post $\dfrac{\log{36}}{\log{6}}$ ? $\log_b(3x)=25 \implies \log_b(3) + \log_b(x) = 25 \implies \log_b(3)=25-\log_b(x)$ Multiply every term in the last equation by 3 ... $3\log_b(3) = 75 - 3\log_b(x)$ compare to your last equation $\log_b(27)=k-3\log_b(x)$ ... what is $k$?
Ah yes sorry was half asleep it is Log 36. Thanks for helping with that last question. How you even figured that out is insane and how you went from there. Genius. I wouldn't have thought that you were suppose to make the first equation look like the 2nd. I was thinking of it in a completely different way. Does that first question make any sense to you? I know it's rather rambly.

Last edited by zekea; December 15th, 2016 at 06:08 AM.

 December 15th, 2016, 06:11 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983 $\dfrac{\log{36}}{\log{6}}=\dfrac{\log{6^2}}{\log{ 6}}=\dfrac{2\log{6}}{\log{6}}=\dfrac{2\cancel{\log {6}}}{\cancel{\log{6}}}=2$ Thanks from zekea
December 15th, 2016, 06:24 AM   #5
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 Originally Posted by skeeter $\dfrac{\log{36}}{\log{6}}=\dfrac{\log{6^2}}{\log{ 6}}=\dfrac{2\log{6}}{\log{6}}=\dfrac{2\cancel{\log {6}}}{\cancel{\log{6}}}=2$
Right, I'm following that and that's what I did to answer the question, but I'm confused as to why I can bring down the 2. I get 36 becomes 6^2 but I'm confused as to why I can bring down the 2 and it's not an exponent anymore, or is it still an exponent number? I understand it's a power law I believe it's called but I"m a bit confused how that works. When you bring down the 2 as the answer, what does the 2 represent? Is it still an exponent value?

Could you rewrite that in terms of an exponential function instead of logs?

Last edited by zekea; December 15th, 2016 at 06:31 AM.

 December 15th, 2016, 07:16 AM #6 Math Team   Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983 I assume you are familiar with the product property of logs? That is ... $\log(a\cdot b)= \log{a}+\log{b}$ note ... $\log(a^2)=\log(a \cdot a)=\log{a}+\log{a}=2\log{a}$ In general ... $\log(a^n) = \log(\text{n factors of a}) = \log{a}+\log{a} + ... +\log{a} \text{ (n terms of log a) }= n\log{a}$
 December 15th, 2016, 10:12 AM #7 Math Team   Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983 prefer this format for a description (had to reference Latex commands) ... $\log(a^n)=\log(\underbrace{a \cdot a \cdot a \cdot ... \cdot a}_\text{n factors}) = \underbrace{\log{a}+\log{a}+\log{a}+...+\log{a}}_{ \text{n terms}} = n\log{a}$ Thanks from zekea
December 15th, 2016, 10:13 AM   #8
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 Originally Posted by skeeter prefer this format for a description (had to reference Latex commands) ... $\log(a^n)=\log(\underbrace{a \cdot a \cdot a \cdot ... \cdot a}_\text{n factors}) = \underbrace{\log{a}+\log{a}+\log{a}+...+\log{a}}_{ \text{n terms}} = n\log{a}$
ees beautiful!

December 15th, 2016, 02:31 PM   #9
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 Originally Posted by skeeter I assume you are familiar with the product property of logs? That is ... $\log(a\cdot b)= \log{a}+\log{b}$ note ... $\log(a^2)=\log(a \cdot a)=\log{a}+\log{a}=2\log{a}$ In general ... $\log(a^n) = \log(\text{n factors of a}) = \log{a}+\log{a} + ... +\log{a} \text{ (n terms of log a) }= n\log{a}$
So the squared a is squaring the answer of X^N=a correct? So in exp form this would be 10^n = A^2. Is that correct?
Also how I write log36/log6 and solve that in exp form instead of log form?

Last edited by zekea; December 15th, 2016 at 02:42 PM.

December 15th, 2016, 05:15 PM   #10
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 So the squared a is squaring the answer of X^N=a correct? So in exp form this would be 10^n = A^2. Is that correct?
This statement makes no sense at all ... where do you get the idea that $10^n = a^2$ ?

Note that $y = \log_{10}(a^n) \implies 10^y = a^n$

The power property of logarithms states $\log_b(a^n) = n \log_b(a)$, which is a generalization for evaluating the log of any value raised to any power, which is what $n$ represents.

Have you been introduced to all properties of logarithms and how they are derived?

1. $\log_b(b^z) = z$ ... definition of a logarithm, $b^z = b^z$

2. $\log_b(1) = 0$ because $b^0 = 1$

3. $\log_b(xy) = \log_b(x)+\log_b(y)$ ...

If $p = \log_b(x)$ then $x = b^p$ and if $q = \log_b(y)$ then $y = b^q$.

Therefore ... $xy = b^p \cdot b^q = b^{p+q} \implies \log_b(xy) = \log_b(b^{p+q}) = p+q = \log_b(x)+\log_b(y)$

4. 3. $\log_b(x/y) = \log_b(x)-\log_b(y)$ ...

If $p = \log_b(x)$ then $x = b^p$ and if $q = \log_b(y)$ then $y = b^q$.

Therefore ... $x/y = b^p/b^q = b^{p-q} \implies \log_b(x/y) = \log_b(b^{p=q}) = p-q = \log_b(x)-\log_b(y)$

5. $\log_b(x^n) = n\log_b(x)$ ... proven in posts #5 & #6.

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