Logarithm Equation Question 3

This is an interesting question which I came across under Additional Mathematics (A-Math):

Find the value of x.

$(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}$

Who is courageous to work on this question?

I will respond to this question when someone discusses about this question first :)

Update:

$(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}$

(Apply Power Law)

$(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}$

(multiply by 4 on both sides of equal sign)

$(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}$

(Apply Power Law)
$(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}$

$(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}$

(Apply Quotient Law)
$(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}$
$(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}$

(Factorise)

$(log_x \sqrt{3})(log_x\sqrt{8})= \frac{3}{2}log_x\frac{1}{\sqrt{2}}$

$(log_x8)=0 (NA)$ $or log_x3+1=0$

$log_x3=-1,3=x^-^1,3=\frac {1}{x},3x=1,x= \frac {1}{3}$

Ai Ling Ong

Hi, I'm Ai Ling Ong. I enjoy coaching students who have challenges with understanding and scoring in 'O' Level A-Maths and E-Maths. I develop Math strategies, sometimes ridiculous ideas to help students in understanding abstract concepts the fast and memorable way. I write this blog to share with you the stuff I teach in my class, the common mistakes my students made, the 'way' to think, analyze... If you have found this blog post useful, please share it with your friends. I will really appreciate it! :)

Comments

AGK says

April 13, 2008 at 12:11 am

(log-baseX-root3)(log-baseX-root8)=1.5(log-baseX-1/root2)

(0.5log-baseX-3)(1.5log-baseX-2)=1.5(log-baseX-2^-0.5)

[cancel 1.5 from both sides]

(0.5log-baseX-3)(log-baseX-2)= -0.5(log-baseX-2)

[cancel 0.5 from both sides]

(log-baseX-3)(log-baseX-2) + (log-baseX-2)= 0

(log-baseX-2) (log-baseX-3 + 1) = 0

therefore,

(log-baseX-2)=0 or (log-baseX-3 + 1)=0

X^0= 2 (rejected) (log-baseX-3)= -1
X^(-1) = 3
thus X= 1/3

hope i didnt make any mistakes
Logarithm rocks! XDD
alwaysLovely says

May 8, 2008 at 7:30 pm

awesome!
Tks for the effort.
ahm97sic says

November 8, 2008 at 1:06 am

This question is fun. Perhaps, you will like to let your students to try to solve the question.

Solve the equation

(2/x)^(log n 4) - (3/x)^(log n 9) = 0

Regards,

ahm97sic
ahm97sic says

November 12, 2008 at 10:37 am

This is another interesting question. Perhaps, you will like to let your students to try to solve the question.

Solve (log a X)^(log b X) = X

where a, b are positive real numbers except 1, leave

the answer in terms of a and b.

Regards,

ahm97sic
Malakia chris says

May 11, 2011 at 5:12 pm

Hie you are just an amazing lady thanks for everything.i will chat with you next time,tell me about your place the temperature & the environment

Ai Ling Ong

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