O Level Logarithm Question (Challenging)

Question:

Given $\displaystyle\log_9{a} = \log_{12}{b} =\log_{16}{(a+b)}$ , find the value of $\displaystyle\frac{a}{b}$ .

Solution:

Working with logarithm is tricky, we try to transform the question to an exponential question.

Let $\displaystyle y=\log_9{a} = \log_{12}{b} =\log_{16}{(a+b)}$

Then, we have $a=9^y=3^{2y}$

$b=12^y=3^y\cdot 2^{2y}$

$a+b=16^y=2^{4y}$ .

Here comes the critical observation:

Observe that $\boxed{a(a+b)=b^2}$ .

Divide throughout by $b^2$ , we get $\displaystyle (\frac{a}{b})^2+\frac{a}{b}=1$ .

Hence, $\displaystyle (\frac{a}{b})^2+\frac{a}{b}-1=0$ .

Solving using quadratic formula (and reject the negative value since $a$ and $b$ has to be positive for their logarithm to exist),

We get $\displaystyle\frac{a}{b}=\frac{-1+\sqrt{5}}{2}$ .

If you have any questions, please feel free to ask me by posting a comment, or emailing me.

(I will usually explain in much more detail if I teach in person, than when I type the solution)

Author: mathtuition88

Math and Education Blog View all posts by mathtuition88

7 thoughts on “O Level Logarithm Question (Challenging)”

tomcircle says:

May 12, 2013 at 12:57 am

Well done !
Good observation a(a+b) = b^2 which is the key leads to solving quadratic equation in (a/b).

LikeLike

Reply
Pingback: Singapore Math (High School): Logarithm Question | Singapore Maths Tuition
JOSEPH says:

December 17, 2016 at 3:59 am

I have liked the identity a(a+b)=b2. explain more

LikeLiked by 1 person

Reply
1. mathtuition88 says:
  
  December 17, 2016 at 11:55 am
  
  Note that both $a(a+b)$ and $b^2$ are equal to $3^{2y}\cdot 2^{4y}$ !
  
  LikeLike
  
  Reply
Pingback: Free Logarithm Worksheet for Secondary School | Singapore Maths Tuition
iamvistinginstructoreric says:

September 9, 2019 at 3:37 pm

Reblogged this on Project ENGAGE.

LikeLike

Reply
Pingback: Hardest Questions in Additional Mathematics (A Math) – Singapore Maths Tuition

Leave a Reply Cancel reply

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: