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November 21st, 2017, 07:39 AM  #1 
Newbie Joined: Oct 2015 From: Greece Posts: 28 Thanks: 0  Logarithm Calculations Help
I'm a university student and still can't do calculations with logarithms i'm trying to calculate this: log8(4) <=> 8^x = 4 but i can't find x!!! I mean how can u multiply 8, x times with itself and find 4!!! This is insane!!! This made me think something that i never thought about all these years. I've seen a lot of times that we can have real numbers as power and i've used them a lot but i never tried to do a calculation like this: 8 ^ (1/2) = ?, how am i going to multiply 8, 1/2 times with itself? This seems so weird!!! Please i must understand all of these else i will never get my master 
November 21st, 2017, 07:48 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,678 Thanks: 1339 
change the base to 2 ... $\log_8{4} = \dfrac{\log_2{4}}{\log_2{8}} = \dfrac{2}{3}$ $8^{1/2} = \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}$ 
November 21st, 2017, 08:02 AM  #3 
Newbie Joined: Oct 2015 From: Greece Posts: 28 Thanks: 0  
November 21st, 2017, 08:08 AM  #4 
Newbie Joined: Oct 2015 From: Greece Posts: 28 Thanks: 0 
So i searched about your answer and i found this formula. d can be any number i want? And also, why do i have to change the base in order to find the logarithm of 4 with base 8? 
November 21st, 2017, 08:27 AM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,678 Thanks: 1339  Quote:
so, using the same method, what base number would you use to determine the value of $\log_9{27}$ ?  
November 21st, 2017, 08:39 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,678 Thanks: 1339 
Note that you could also determine $\log_8{4}$ using exponents ... $\log_8{4} = x$ $8^x = 4$ $(2^3)^x = 2^2$ $2^{3x} = 2^2 \implies 3x = 2 \implies x = \dfrac{2}{3}$ ... either way, one must recognize a common base value. 
November 21st, 2017, 08:41 AM  #7  
Newbie Joined: Oct 2015 From: Greece Posts: 28 Thanks: 0  Quote:
I'm really grateful!  
November 21st, 2017, 09:32 AM  #8 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,661 Thanks: 965 Math Focus: Elementary mathematics and beyond 
$$\log_927=3\log_93=3\cdot\frac12=\frac32$$

November 21st, 2017, 12:35 PM  #9 
Senior Member Joined: May 2016 From: USA Posts: 857 Thanks: 348 
It is possible to define rational exponents in terms of repeated multiplications of a base or a root of a base. But exponents that are irrational numbers can be defined only in terms of limits.


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calculations, logarithm 
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