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indices and logarithms some question Rate Topic: -----

#1 13 February 2012 - 10:15 AM Vastor 


Baryon
hey guys,

well, as we know, adding among log like this is easy

\log_c A + \log_c B = \log_c AB

but how does to "add" something like this?

\log_3 12-3^x = x - 1

and basically, how you calculate add in an equation that has indices and logarithms?

I mean if

A * A = A^{1+1}

but how about
A + A = A^? = 2A
Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.
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#2 13 February 2012 - 10:54 AM ajb 


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Physics Expert
Equalities like

x = \log_{a}(x^{a}) = a^{\log_{a}(x)},

maybe useful.

______________________

There us an obvious typo here!

x = \log_{a}(a^{x})

This post has been edited by ajb: 15 February 2012 - 10:55 AM

"In physics you don't have to go around making trouble for yourself - nature does it for you" Frank Wilczek.

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#3 13 February 2012 - 01:37 PM Fuzzwood 


Formerly known as Fswd
Another hint: 12 = 9 + 3
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#4 13 February 2012 - 02:50 PM Vastor 


Baryon
ouch, I don't know what the cause, my brain hurt so hard when dealing with logs that involve adding like this question, indices is much better alternative

3^{x-1} + 3^x = 12 (originals of above question)

using factorisation and other stuff without involving log can be solved easily and get the answer of x = 2.

but can log solve above problem also? I try but I had problem when I'm trying to convert the indices to a log because of the add/minus symbol. it give me so much headache on how to create a symbol for the add/minus (like multiply and divide they turn into plus/minus in log) if then, what plus/minus turn out to be?

even using ajb recommendation, I still can't convert it in which I'm sure I'm do it right

if x = \log_{a}(x^{a}) = a^{\log_{a}(x)}

then 3^{x-1} + 3^x = \log_{1}{12^1}
i'm doing it right? if then, how I convert on the left part to log? arrrghh :wacko:
Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.
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#5 14 February 2012 - 08:31 AM Vastor 


Baryon
anyone?:mellow:
Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.
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#6 14 February 2012 - 11:02 AM ajb 


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Physics Expert
Okay. A good hint would be to multiply both sides of your equation by 3^{1-x}. Use the law of indices and then take your logs.
"In physics you don't have to go around making trouble for yourself - nature does it for you" Frank Wilczek.

My homepage.
1

#7 15 February 2012 - 08:05 AM Vastor 


Baryon

View Postajb, on 14 February 2012 - 11:02 AM, said:

Okay. A good hint would be to multiply both sides of your equation by 3^{1-x}. Use the law of indices and then take your logs.

ok, I got it, thnx, so the clue here is to put the variable x in one place only(aka not seperated between two number by the plus symbol)?!
Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.
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#8 15 February 2012 - 08:20 AM ajb 


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Physics Expert
I think so.

One is tempted to write

\log(a+b) = \log(a)+ log(b),

but this is not true. I cannot see any more direct way of using logs in tackling this question. Though note everything is quite easy in your example and the answer can just be spotted.
"In physics you don't have to go around making trouble for yourself - nature does it for you" Frank Wilczek.

My homepage.
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#9 15 February 2012 - 09:50 AM Vastor 


Baryon

View PostVastor, on 15 February 2012 - 08:05 AM, said:

ok, I got it, thnx, so the clue here is to put the variable x in one place only(aka not seperated between two number by the plus symbol)?!

forget this, this one is clue on how to solve the question's problem...

View Postajb, on 15 February 2012 - 08:20 AM, said:

I think so.

One is tempted to write

\log(a+b) = \log(a)+ log(b),

but this is not true. I cannot see any more direct way of using logs in tackling this question. Though note everything is quite easy in your example and the answer can just be spotted.


I understand that, when first seeing log, I can't really see the pattern on how it work, and can't even understand why

A^B = C
become
log_A C = B

until now, the only thing not really clear to me is what does A represent (which why give me so much difficulty on converting from indice to log or otherwise), yes it is "base" but what is it actually, does it refer to the same "base" of "number type"(sorry, not really know many definition on math here), I mean does it refer to base-ten / base-two number?

if so, why does
log_1 12^1 = 12
does not 12 is number base-10? why it is 1 in log?

for the part that I understand on log:-
- they behave like indices, except they are being as the "main number". example:-
10^5 = 100,000
become
log 100,000 = 5 the 5 became the "main number", and the "original main number" become inner-side of the log.

- we can change the number base(and the "original main number") as we like without changing its "main number" values. examples:-
1 = log 10 = log_5 5 = log_6 6 = 1

- there some pattern on how there some relation between the "base" and "main number" and "inner-side number"(all number, nuff said)in log
log_9 \frac{x}{y}

=\frac{log_3 \frac{x}{y}}{log_3 9}

=\frac{log_3 \frac{x}{y}}{2}

=\frac{log_3 x - log_3 y}{2}

=\frac{1}{2}(log_3 x - log_3 y)

=log_3 \sqrt{x} - log_3 \sqrt{y}

can you say something about this?(wikipedia just not enough):lol:

edit: should proof-read first

This post has been edited by Vastor: 15 February 2012 - 09:52 AM

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the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.
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#10 15 February 2012 - 11:01 AM ajb 


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Physics Expert
There is a typo earlier on, should have spotted that before.

The definition of of the log is

x= \log_{a}(a^{x}).

Now, log to the base one is not defined. One to any power is still one.
"In physics you don't have to go around making trouble for yourself - nature does it for you" Frank Wilczek.

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#11 17 February 2012 - 02:25 AM Vastor 


Baryon
edit : Ignore this post, should have read question better :D

hey guys,

I don't know if the problem is in the answer provided in the book but the exercise said

Solve the following equation 9log_3 x = 7

9log_3 x = 7 


 log_3 x = \frac{7}{9} 



 3^{\frac{7}{9}} = x 



answer in the book

 x = \sqrt{7}



ooo wait, does it legitimate to use graphing calculator like this to find the answer?

Posted Image

lost my calculator somewhere :doh:
						
						

						

								This post has been edited by Vastor: 17 February 2012 - 07:52 AM
								
							

					

					
						

	Some sort of good tutor :}



the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.


					
				


			

				
												
				
					
				
				
				
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						#12
						
						
						17 February 2012 - 07:52 AM
						


						ajb 
						
					



				





	
 
	

			
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Physics Expert
	
	


					
					
					

						Your answer x= 3^{\frac{7}{9}} is correct.
						
						

						
					

					
						

	"In physics you don't have to go around making trouble for yourself - nature does it for you"   Frank Wilczek.



My homepage.


					
				


			

				
												
				
					
				
				
				
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						#13
						
						
						18 February 2012 - 06:51 AM
						


						Vastor 
						
					



				





	
 
	
	
			
Baryon
	
	


					
					
					

						Posted Image



how can this happened?
						
						

						
					

					
						

	Some sort of good tutor :}



the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.


					
				


			

				
												
				
					
				
				
				
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						#14
						
						
						19 February 2012 - 09:17 AM
						


						Vastor 
						
					



				





	
 
	
	
			
Baryon
	
	


					
					
					

						hey guys,

ignore my post before and never use your pc calculator to calculate log xD



anyway, I got another sort of hard question:



Solve the equation  4^x - 7*2^x - 8 = 0



  4^x - 7*2^x - 8 = 0 



  4^x - 7*2^x = 8 



 2^x(2^x - 7) = 2^3 



 2^x - 7 = \frac{2^3}{2^x} 



using trial and error, you can get x = 3



but how to calculate it?
						
						

						

								This post has been edited by Vastor: 19 February 2012 - 09:19 AM
								
							

					

					
						

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the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.


					
				


			

				
												
				
					
				
				
				
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						#15
						
						
						19 February 2012 - 12:18 PM
						


						Fuzzwood 
						
					



				





	
 
	
	
			
Formerly known as Fswd
	
	


					
					
					

						You could pull 7 apart in 23 - 20. Then you have everything as 2 to the power of something, and something to the power of 0 which always equals 1:



x - (3 - 1) = 3/x



x - 3 + 1 = 3/x



x - 2 = 3/x



x² - 2x - 3 = 0



(x - 3)(x + 1)



x = 3 or x = -1
						
						

						
					

					
				


			

				
												
				
					
				
				
				
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						#16
						
						
						19 February 2012 - 12:29 PM
						


						Vastor 
						
					



				





	
 
	
	
			
Baryon
	
	


					
					
					

						
View PostFuzzwood, on 19 February 2012 - 12:18 PM, said:

You could pull 7 apart in 23 - 20. Then you have everything as 2 to the power of something, and something to the power of 0 which always equals 1:



x - (3 - 1) = 3/x



x - 3 + 1 = 3/x



x - 2 = 3/x



x² - 2x - 3 = 0



(x - 3)(x + 1)



x = 3 or x = -1




don't you think there error in your calculation?



how do you get



x - (3-1) = 3/x ?
						
						

						
					

					
						

	Some sort of good tutor :}



the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.


					
				


			

				
												
				
					
				
				
				
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						#17
						
						
						22 February 2012 - 02:14 PM
						


						imatfaal 
						
					



				





	
 
	

			
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Primate
	
	


					
					
					

						To check answers when unsure always sub back into the original equation



  4^x - 7*2^x - 8 = 0 



  4^3 - 7*2^3 - 8 = 0 



 64 - 7*8 - 8 = 0 



 64 - 56 - 8 = 0 



ok that's cool



  4^{-1} - 7*2^{-1} - 8 = 0 



 1/4 - 7*(1/2) - 8 = 0 



1/4  - 7/2 - 8 = 0 



and that is not right!

the error in Fswd's answer is in the taking of logs of multiplied factors



I would do it this way



  4^x - 7*2^x - 8 = 0  



 2^{2x} - 2^x(2^3-2^0) - 2^3 = 0   



 2^{2x} - 2^x.2^3+2^x.2^0 - 2^3 = 0   



and bracketed to make the correct removal of the 2s clearer



 2^{2x} - (2^x.2^3)+(2^x.2^0) - 2^3 = 0  

 

 2x   		- (x+3)+(x+0) - 3 = 0  



 2x   		- x-3+x - 3 = 0  



 2x   		-3- 3 = 0  



 2x   		= 6 



 x=3  
						
						

						
					

					
						

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						#18
						
						
						27 April 2012 - 12:14 AM
						


						Vastor 
						
					



				





	
 
	
	
			
Baryon
	
	


					
					
					

						Hey guys,

35. It is known that x and y are related by the equation  y = ax^n , where a and n are constants. When a graph of  log_2 y  against  log_2 x  is plotted, a straight line passing through the points (1,5) and (3,11) is obtained. Find the values of a and n.



 y = ax^n \Rightarrow lg y = lg a + n*lg x \Rightarrow log_2 y = n*log_2 x + log_2 a, right?



 log_2 y = m*log_2 x + C, P_1(3, 11), P_2(1, 5)



 m = \frac{11-5}{3-1} = \frac{6}{2} = 3  (real answer = 3)



if x = 1, y = 5.

 log_2 5 = 3*log_2 1 + C \Rightarrow lg 5 = lg 1 + C \Rightarrow lg 5 = C 



where C = lg a

 lg 5 = lg a \Rightarrow a = 5  (real answer = 4)





if x = 3, y = 11.

 log_2 11 = 3*log_2 3 + C \Rightarrow lg 11 = 3*lg 3 + C \Rightarrow C = lg 11 - lg 27 \neq lg 5  how this happen? does Linear Law breaks down its "accurate" relation? or just my daily 'mathypo'.
						
						

						
					

					
						

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