My Math Forum Logarithm Calculations Help

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 November 21st, 2017, 06:39 AM #1 Member   Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6 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, 06:48 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,771 Thanks: 1426 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, 07:02 AM   #3
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Quote:
 Originally Posted by skeeter change the base to 2 ... $\log_8{4} = \dfrac{\log_2{4}}{\log_2{8}} = \dfrac{2}{3}$ $8^{1/2} = \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}$
About the second answer, i feel ashamed I think i "burned" my mind.
But the first one didn't know it.

 November 21st, 2017, 07:08 AM #4 Member   Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6 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, 07:27 AM   #5
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Quote:
 Originally Posted by babaliaris 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?
yes, d can be any positive number you wish ... I chose 2 as the base because both values can be expressed as powers of 2, i.e $2^2 = 4$ and $2^3 = 8$

so, using the same method, what base number would you use to determine the value of $\log_9{27}$ ?

 November 21st, 2017, 07:39 AM #6 Math Team   Joined: Jul 2011 From: Texas Posts: 2,771 Thanks: 1426 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, 07:41 AM   #7
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Quote:
 Originally Posted by skeeter yes, d can be any positive number you wish ... I chose 2 as the base because both values can be expressed as powers of 2, i.e $2^2 = 4$ and $2^3 = 8$ so, using the same method, what base number would you use to determine the value of $\log_9{27}$ ?
log3(27) / log3(9) = 3/2

I'm really grateful!

 November 21st, 2017, 08:32 AM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond $$\log_927=3\log_93=3\cdot\frac12=\frac32$$
 November 21st, 2017, 11:35 AM #9 Senior Member   Joined: May 2016 From: USA Posts: 1,150 Thanks: 479 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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