Changing the base of a logarithm

Question

I must simplify $\log_4 (9) + \log_2 (3)$. I have tried but I can't get the correct answer $2 \log_2 (3)$. How do I proceed?

Answer 1

We can use $\log_b x+\log_b y=\log_b xy$ to get $\log_v 23+\log_v 49=\log_v 1127$. However, using $\log_b x^y=y\log_b x$, we see that the "correct answer" is equal to $\log_v 23^2=\log_v 529\neq \log_v 1127$

Answer 2

It is known, that $log_{a^2}(b^2)=log_a(b)=x$. You can write it in an exponential form: $(a^2)^x=b^2$ and $a^x=b$. Both equations have the same solution.

In your case it is $log_4(9)+log_2(3)=log_2(3)+log_2(3)=2log_2(3)$

Answer 3

The good way is to use calculus's answer.

Another way (I am lazy and I hate bases !) is to convert everything to natural logarithms. So, $$ \log_2 (3)+\log_4 (9) =\frac{\log (3)}{\log (2)}+\frac{\log (9)}{\log (4)}=\frac{\log (3)}{\log (2)}+\frac{\log (3^2)}{\log (2^2)}=\frac{\log (3)}{\log (2)}+\frac{2\log (3)}{2\log (2)}=2\frac{\log (3)}{\log (2)}$$ $$ \log_2 (3)+\log_4 (9) =2\log_2 (3)$$