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Gian

algebraic headache

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Excuse my ignorance, but with an equation like this; 

32 x 0.5= 1

is there a formula I can use to calculate the index number x ?

cheerz

GIAN :)

Edited by Gian

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You do not need any formulas. How about you think about what you are asked: how many times do you have to multiply .5 = 1/2 to itself to get 1/32? How would that work? Like 1/2 x 1/2 = 1/4 does not quite get there, neither does 1/2 x 1/2 x 1/2 = 1/8.  You need to multiply a 2 together with itself enough times to get 32 as the answer.

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7 hours ago, Gian said:

32 x 0.5= 1

is there a formula I can use to calculate the index number x ?

The more appropriate mathematical naming conventions: x is called exponent.

https://en.wikipedia.org/wiki/Exponentiation

 

32 * (1/2)^x = 1

(1/2)^x = 1/32

2^-x = 1/32

2^-x = 32^-1

2^x = 32

2^5 = 32

log2(32)=5

 

 

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9 hours ago, Sensei said:

The more appropriate mathematical naming conventions: x is called exponent.

https://en.wikipedia.org/wiki/Exponentiation

 

32 * (1/2)^x = 1

(1/2)^x = 1/32

2^-x = 1/32

2^-x = 32^-1

2^x = 32

2^5 = 32

log2(32)=5

 

 

THANKS!!!☺☺☺

Edited by Gian

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On 26/06/2018 at 8:09 PM, taeto said:

You do not need any formulas. How about you think about what you are asked: how many times do you have to multiply .5 = 1/2 to itself to get 1/32? How would that work? Like 1/2 x 1/2 = 1/4 does not quite get there, neither does 1/2 x 1/2 x 1/2 = 1/8.  You need to multiply a 2 together with itself enough times to get 32 as the answer.

THANKS!!!!! Got it (I think)

1/2x1/2=1/4 (viz to the power2)

x1/2= 1/8            (to the power3)

x1/2= 1/16          (to the power4)

x1/2= 1/32          (to the power5)      So ans=5

Is there a particular formula I should use though? Or is working through it methodically number by number the right way to do it? :)

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4 hours ago, Gian said:

x1/2= 1/32          (to the power5)      So ans=5

Is there a particular formula I should use though? Or is working through it methodically number by number the right way to do it? :)

The engineering way to do it is to say: if you have your number x = 1/2 and your other number y = 1/32, and you wonder how to compute a number a for which x^a = y, then you just have to grab your calculator and punch in (log y)/(log x) to get the answer (approximately). 

That skips some amount of mathematics which is involved. Especially if you are not familiar with how logarithms work. Logarithms have the property that if you want to compute a product xz of two numbers x and z, but the multiplication looks hard, then you can instead compute log(x) and log(z), and then the sum y = log(x) + log(z) will have the property log(xz) = y, so you can deduce the answer by looking for a number w such that log(w) = y. It seems a roundabout method, but some people can do this very routinely by the use of slide rules. A slide rule will allow you to find log(x) and log(z) quickly, and to add their values, and it allows you to quickly read off the value of w as well.

More generally, logarithms have the property log(x^n) = n log(x), i.e, if you multiply x by itself n times, then the logarithm of the answer has to be what you get by adding the log(x) to itself n times. So if x^n = y, then n log(x) = log (y), and you immediately get n = log(y)/log(x).

 

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Some you could work by hand, but that isn't often going to be practical. Logarithms are your friends in dealing with exponents.

ex.

32 * 0.5x = 1

32/32 * 0.5x = 1/32

log0.5(0.5x) = log0.5(1/32)

x = log0.5(1/32)

x = 5

 

cross posted with taeto

Edited by Endy0816

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