My Math Forum A couple Logarithm Questions

 Pre-Calculus Pre-Calculus Math Forum

December 16th, 2016, 02:46 AM   #11
Newbie

Joined: Dec 2016

Posts: 12
Thanks: 0

Quote:
 Originally Posted by skeeter 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.
Ok sorry yes, that's what I was trying to articulate. $y = \log_{10}(a^n) \implies 10^y = a^n$ So you're raising whatever your answer is from 10^y to some sort of power. Thanks for writing out those log laws I've taken some time to look at them.

I have one more question to bother you with. This is a question in a worksheet.

The equation m log p (n) = q can be written in exponential form as...

The answer on the work sheet is p^(q/m)=n but shouldn't it be P^(qm) = n ? According to the power rule? My teacher explained this by writing down for me log p (n) = q / m but I'm confused here.

I got my answer from Khans' video here

Basically according to the power rule you have Log a (c)^d = bd . He brought the d down to the other side.
So in exp form A^(bd) = C^d so shouldn't the answer be p^(mn) = n rather than P^(q/m) = n ?

I get the feeling that what Khan is doing is different to what the question is asking. Khan is proving a law whereas as this is manipulating a log equation to exponential form but I'm confused as to how they're different.

 December 16th, 2016, 03:49 AM #12 Math Team   Joined: Jul 2011 From: Texas Posts: 2,674 Thanks: 1337 Two ways to do this ... 1. $m\log_p(n)=q$ divide both sides by $m$ ... $\log_p(n)=\dfrac{q}{m}$ change to an exponential ... $n=p^{q/m}$ 2. $m\log_p(n)=q$ rewrite using the power property for logs ... $\log_p(n^m)=q$ change to an exponential ... $n^m=p^q$ Both results are the same. Note ... $n=p^{q/m}$ raise both sides to the $m$ power ... $n^m = \left(p^{q/m}\right)^m = p^{(q/m) \cdot m} = p^q$
 December 16th, 2016, 04:23 AM #13 Newbie   Joined: Dec 2016 From: Canada Posts: 12 Thanks: 0 Okay that makes sense to me but why does Khan in the video do then log a (c)^d = bd . The D is being multiplied rather than divided. Which in exp form is A^(bd) = c^d. Is he talking about something else?
 December 16th, 2016, 07:56 AM #14 Math Team   Joined: Jul 2011 From: Texas Posts: 2,674 Thanks: 1337 I didn't watch the video ...

 Tags couple, logarithm, questions

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post mike1127 Pre-Calculus 4 September 5th, 2015 12:49 PM Drasik Geometry 11 October 10th, 2014 11:57 AM Scorks Calculus 1 May 6th, 2012 08:46 PM Valar30 Calculus 2 November 15th, 2010 09:21 AM bignick79 Algebra 3 June 15th, 2010 06:44 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top