A.Maths Qn (Indices & Log)

wishboy SG

this is a difficult qn (to me) i encountered when doing past yr papers
can someone pls help me with this?

Solve for x.

skythewood SG

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.

Ahm97sic SG

Hi Wishboy,

This question is fun ie it makes use of 2 interesting (but not so commonly used)concepts to solve the question.

The answer is x = 6.

Thank you or your kind attention.

Regards,

ahm97sic

wishboy SG

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?

skythewood SG

i see you change the log from 4 to 2, 9 to 3.

that got me thinking... possible to use A^2 - B^2 = (A+ B)(A-B) ?

than you will have 2 answers

skythewood SG

no idea

wishboy SG

Ahm97sic, can u help me continue? T.T

skythewood SG

Ahm97sic, can u help me continue? T.T

can change log n 2 = (log 10 2)/(log 10 n) ? is this legal change of base?

wishboy SG

ya can change like tat
but it is still a power, only way to bring down is to make the base the same as the number

skythewood SG

last resort:

bring the power from the x side to the number side,

change on base on the number side.

not sure if this is legal though

Ahm97sic SG

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

Mikethm SG

Mathtype expressions can be saved as gif i mean.

Ahm97sic SG

Hi Mikethm,

Thank you for the information.

Regards,

ahm97sic

wishboy SG

ok i have received the file

so the main step is to bring over the power

thanks everyone for ur help

Ahm97sic SG

Hi Wishboy,

I have sent the answer in an email with the attached file to your email account.

I like this question. It is fun. Please post more.

Thank you for your kind attention.

Regards,

ahm97sic

Uncertain SG

Equate (2/x) ^ log = (3/x) ^ log

Then take log base n on both side.

Simplify the equations to arrive at log n 6 (log n 2/3) = log n x (log n 2/3)

Ans: x = 6

Mega wolf005 SG

Can you send me the answer too cause i really don't get how to do it, my email is [email protected] thank you very much

secretliker SG

(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

Mikethm SG

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?)