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November 21st, 2017, 06:39 AM   #1
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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
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November 21st, 2017, 06:48 AM   #2
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change the base to 2 ...

$\log_8{4} = \dfrac{\log_2{4}}{\log_2{8}} = \dfrac{2}{3}$


$8^{1/2} = \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}$
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November 21st, 2017, 07:02 AM   #3
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Quote:
Originally Posted by skeeter View Post
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.
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November 21st, 2017, 07:08 AM   #4
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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?
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November 21st, 2017, 07:27 AM   #5
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Quote:
Originally Posted by babaliaris View Post
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 $\log_9{27}$ ?
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November 21st, 2017, 07:39 AM   #6
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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.
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November 21st, 2017, 07:41 AM   #7
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Quote:
Originally Posted by skeeter View Post
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!
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November 21st, 2017, 08:32 AM   #8
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$$\log_927=3\log_93=3\cdot\frac12=\frac32$$
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November 21st, 2017, 11:35 AM   #9
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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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