challenging – Singapore Maths Tuition

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)

Tag: challenging