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)