# 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

https://mathtuition88.com/

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

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

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2. JOSEPH says:

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

Liked by 1 person

1. Note that both $a(a+b)$ and $b^2$ are equal to $3^{2y}\cdot 2^{4y}$!

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