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May 29, 2024, 04:00:17 pm
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Author Topic: help  (Read 746 times)  Share 

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Teen

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help
« on: July 14, 2012, 10:52:19 pm »
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guys need help with this question
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Kaille

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Re: help
« Reply #1 on: July 14, 2012, 11:07:03 pm »
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just wondering but do you have the answers? i just want to check whether i've done something completely dumb before putting it up my working out and making a fool of myself...:P
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Teen

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Re: help
« Reply #2 on: July 14, 2012, 11:18:13 pm »
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Quote from: Kaille on July 14, 2012, 11:07:03 pm
just wondering but do you have the answers? i just want to check whether i've done something completely dumb before putting it up my working out and making a fool of myself...:P

i don't have the answer. can pls put ur working out. it would be helpful
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b^3

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Re: help
« Reply #3 on: July 14, 2012, 11:23:58 pm »
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You may want to revise your log laws/rules. I'm not going to show all working as you should have a go at it with what I've said first.
1) You can sub in the to get , then you can bring the power of 2 down to the front (refer to log laws), resulting in .
2) Let , then convert it to an exponential, i.e. . Then take the log with the base of both sides, and again refering to the log laws, . Then you should get , from there solve for and use your answer from the first part for . You should get .

Hope that helps.

Also please don't post the same topic twice in two different boards.
« Last Edit: July 14, 2012, 11:25:30 pm by b^3 »
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Kaille

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Re: help
« Reply #4 on: July 14, 2012, 11:24:25 pm »
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ok i'm not sure how to write all the fancy maths stuff, so i'll try and do it here. Be mindful i'm not the most gifted at maths so this could be verrrry wrong....

a. logba=p
c=a^2
logbc= 2logba (i took logb of both sides)
= 2p

b. this one is a change of base question (me thinks)
logcb = logbb / logbc (im not sure if this step is correct)
= 1 / 2p

sorry about the unmathsy setting out, but i hope you can still understand it :P

better get someone to check this tho, don't wanna be giving people the wrong answers!!
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Teen

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Re: help
« Reply #5 on: July 14, 2012, 11:34:27 pm »
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sorry about posting the post twice :). I don't realize that... sorry about that b^3
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