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LoginA.Maths Qn (Indices & Log)
22 posts
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bring one of the term over.
cross multiply.
you will get numbers to the power of the log thing one side,===> change of base
x to the power of the of a log addition subtraction the other side. ===> solve it
maybe you will get the answer. just doing this from visual, no have software to show the maths.
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Originally posted by skythewood:
bring one of the term over.
cross multiply.
you will get numbers to the power of the log thing one side,===> change of base
x to the power of the of a log addition subtraction the other side. ===> solve it
maybe you will get the answer. just doing this from visual, no have software to show the maths.
i got until
mind guiding me on how to continue?
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Hi Wishboy,
I do not know how to type the mathematical symbols eg the log n 4 in the forum.
Do you have any email account that can receive attached file ? I have typed out the answer using Mathtype in Microsoft Word 2003. I can send you an email with the answer in the attached file.
Or I can send the attached file to the moderator Eagle and he will kindly type out the answer in the forum.
Thank you for your kind attention.
Regards,
ahm97sic
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Can you send me the answer too cause i really don't get how to do it, my email is mega_wolf005@hotmail.com thank you very much
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(2/x) ^ logn(4) = (3/x) ^ logn(9)
Square root both sides, not sure if this is legal.
(2/x) ^ logn(2) = (3/x) ^ logn(3)
Change the base at the powers to power 2.
(2/x) ^ [log2(2)/log2(n)] = (3/x) ^ [log2(3)/log2(n)]
Take power log2(n) to remove the denominator.
(2/x) = (3/x) ^ log2(3)
Take log2 on both sides.
1 - log2(x) = log2(3) [log2(3) - log2(x)]
1 - log2(x) = [log2(3)]^2 - log2(3) * log2(x)
1 - [log2(3)]^2 = log2(x) [1 - log2(3)]
Use the identity (A-B)(A+B) = A^2 - B^2
1 + log2(3) = log2(x)
log2(6) = log2(x)
x=6
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Just wondering... why no one want to take the easy way out?
(2/x)^logn4 = (3/x)^logn9
(2/x) = (3/x)^(logn9/logn4)
(2/x) = (3/x)^1.58496
(x^1.58496)/x = (3^1.58496)/2
x^0.58496 = 2.85225
x = 2.85225^(1/0.58496)
x = 6.0000
x = 6.00 (at worse minus 1 mark for unnecessary 3sf but who cares about 1 mark?)
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Oh ok... then you can have this simple use-only-simple-rules-no-calculator solution then :)
(2/x)^2logn2 = (3/x)^2logn3
Taking log to base n on both sides
logn(2/x)^2logn2 = (3/x)^2logn3
2logn2[logn2-lognx] = 2logn3[logn3-lognx]
logn2[logn2-lognx] = logn3[logn3-lognx]
(logn2)^2 -(logn2)(lognx) = (logn3)^2 - (logn3)(lognx)
(logn3)(lognx)-(logn2)(lognx) = (logn3)^2-(logn2)^2
lognx[logn3-logn2] = (logn3-logn2)(logn3+logn2)
lognx = logn3 + logn2
lognx = logn6
x =6
Same method as uncertain's probably. But just for the benefit of those who need it step by step. Does not make use of uncommon concepts. Only make use of 2 simple concepts.1. Must have same base.
2. Making whatever contain x the subject. (E Maths)
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