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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}}

(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}

Logarithm Equation Question 1


I met a student and she passed me this paper with this Lg question on it, asking me on the solution for solving this equation :\lg(x-2)=(lg3)^2

Coincidentally, I was asking a few O Level students of mine on which are the "Killer A-Maths Topics" for them. Logarithm, Indices, Surds is one of the common topic which is ranked high on their list.

Now back to the question.

PS:lg = log_{10}

Notice that right hand side of the equal sign is something you can compute using the calculator hence,

\lg(x-2)=(lg3)^2
(x-2)= 10^{(lg3)^2}

Use your calculator to find out the value of 10^{(lg3)^2} and the value is 1.69 ( round off to 3 significant figure) so the answer for this question is

x = 1.69 + 2<br> =3.69<br>

I will be coming out with a series of posts on Logarithms since it is a "hot" topic for students :-) Subscribe to my feeds to be updated of this post.

Binomial Expansion Teaches how to choose the RIGHT partner (:


I always like to share with my students what each Math topic has to do with their everyday life, particularly their future.

Yesterday, I did Binomial Expansion. It's about finding the RIGHT partner (: It's about mastering the skills of finding the correct match based on a set of factors.

binoqn1.PNG
Just like this question, you see that there are 2 brackets. 1st bracket is good for now. 2nd bracket needs us to do some expansion work by using the FORMULAE (It's provided during GCE O Level Exams but some schools don't seem to provide it for their mid year, strange)
Oh yar, there is NO need to remember the Binomial Expansion Formulae!
binoformulae.PNG
Then one of the questions that often pop up is " When do we stop our expansion for [TEX](1+\frac {x}{3})^{12}[/TEX]?"

To answer this question, it depends on what type of partners you are looking for.

[Read more...] about Binomial Expansion Teaches how to choose the RIGHT partner (: