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C2 Logs... I hate them and need help

(log3(81)+log3(243))/(log3(x)) = log3(x)

Need urgent help with this one.After simplifying the numerator, do I multiply both sides by log3(x) ?
Reply 1
Original post by kieton123
(log3(81)+log3(243))/(log3(x)) = log3(x)

Need urgent help with this one.After simplifying the numerator, do I multiply both sides by log3(x) ?


check my solution
Reply 2
Original post by kieton123
(log3(81)+log3(243))/(log3(x)) = log3(x)

Need urgent help with this one.After simplifying the numerator, do I multiply both sides by log3(x) ?


Yes multiply both sides by log3x

you should end up with log3x squared =9

now square root both sides

log3x= +- 3

solve from there
Reply 3
Original post by kieton123
(log3(81)+log3(243))/(log3(x)) = log3(x)

Need urgent help with this one.After simplifying the numerator, do I multiply both sides by log3(x) ?


3^4 = 81 and 3^5 = 243

therefore your equation is 4+5/log3(x)=log3(x)
let log3(x) be y and make a quadratic equation by multiplying everything by y
i got x as 243 and 1/3.
Reply 4
Original post by RizK
3^4 = 81 and 3^5 = 243

therefore your equation is 4+5/log3(x)=log3(x)
let log3(x) be y and make a quadratic equation by multiplying everything by y
i got x as 243 and 1/3.


Apparently x is 27, Imran got the answer
Reply 5
Original post by kieton123
Apparently x is 27, Imran got the answer


I just checked and this is the way they have done it in the C2 book... mind telling me where you got the question from? :biggrin: thanks
Original post by RizK
3^4 = 81 and 3^5 = 243

therefore your equation is 4+5/log3(x)=log3(x)
let log3(x) be y and make a quadratic equation by multiplying everything by y
i got x as 243 and 1/3.


Answer is x=127, x=27x=\dfrac{1}{27},\ x= 27
log3(81)+log3(243)log3(x)=log3(x)\dfrac{log_{3}(81)+log_{3}(243)}{log_{3}(x)} = log_{3}(x)

log3(81)=4,  log3(243)=5log_{3}(81) = 4, \ \ log_{3}(243) = 5

 9log3(x)=log3(x)\therefore \ \dfrac{9}{log_{3}(x)} = log_{3}(x)

4+5=(log3(x))24+5 = (log_{3}(x))^{2}

log3(x)=9log_{3}(x) = \sqrt{9}

log3(x)=±3log_{3}(x) = \pm 3

[log3(x)=3log3(x)=3]\begin{bmatrix} log_{3}(x) = -3 \\log_{3}(x) = 3 \end{bmatrix}

If loga(x)=blog_{a}(x) = b, then x=abx = a^{b}

 x=33, x=33\therefore \ x = 3^{-3}, \ x = 3^{3}

x=127, x=27x=\dfrac{1}{27}, \ x = 27

You can verify your answer by plugging x back into the original equation and seeing if both sides output the same value.
(edited 7 years ago)
Reply 8
Original post by RizK
I just checked and this is the way they have done it in the C2 book... mind telling me where you got the question from? :biggrin: thanks


Ok man, its the "Revise Edexcel AS Mathematics AS Workbook"
Reply 9
Og my god i read the question wrong xD I thought log3(81)was a serparate terrm lol
Cheers!
Reply 10
sol.PNG
Original post by Reda2
sol.PNG


You're missing a solution > 9=±3\sqrt{9} = \pm 3
You must also consider log3(x)=3log_{3}(x)=-3
Reply 12
Original post by edothero
You're missing a solution > 9=±3\sqrt{9} = \pm 3
You must also consider log3(x)=3log_{3}(x)=-3


Just wanted to show him the solution, pretty sure he knows that.

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