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)=k3LogB (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 $\log_b(27)=k3\log_b(x)$ ... what is $k$? 
December 15th, 2016, 06:01 AM  #3  
Newbie Joined: Dec 2016 From: Canada Posts: 9 Thanks: 0  Quote:
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 
$\dfrac{\log{36}}{\log{6}}=\dfrac{\log{6^2}}{\log{ 6}}=\dfrac{2\log{6}}{\log{6}}=\dfrac{2\cancel{\log {6}}}{\cancel{\log{6}}}=2$

December 15th, 2016, 06:24 AM  #5  
Newbie Joined: Dec 2016 From: Canada Posts: 9 Thanks: 0  Quote:
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) ... $\log(a^n)=\log(\underbrace{a \cdot a \cdot a \cdot ... \cdot a}_\text{n factors}) = \underbrace{\log{a}+\log{a}+\log{a}+...+\log{a}}_{ \text{n terms}} = n\log{a}$ 
December 15th, 2016, 10:13 AM  #8 
Senior Member Joined: Sep 2015 From: CA Posts: 603 Thanks: 324  
December 15th, 2016, 02:31 PM  #9  
Newbie Joined: Dec 2016 From: Canada Posts: 9 Thanks: 0  Quote:
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  
Math Team Joined: Jul 2011 From: Texas Posts: 2,109 Thanks: 983  Quote:
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 ... $x/y = b^p/b^q = b^{pq} \implies \log_b(x/y) = \log_b(b^{p=q}) = pq = \log_b(x)\log_b(y)$ 5. $\log_b(x^n) = n\log_b(x)$ ... proven in posts #5 & #6.  

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