# Changing the base of a logarithm

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?

• Please read this post and format math in the right way. – MonkeyKing May 14 '15 at 23:23
• The correct answer does not look right. – user222031 May 14 '15 at 23:30
• What is your answer? The $2$ outside requires that the original is $\log_v\left(23^2\right)$--which clearly isn't correct. There may be a problem with the answer you think is right. – Jared May 14 '15 at 23:32

Another way (I am lazy and I hate bases !) is to convert everything to natural logarithms. So, $$\log_2 (3)+\log_4 (9) =\frac{\log (3)}{\log (2)}+\frac{\log (9)}{\log (4)}=\frac{\log (3)}{\log (2)}+\frac{\log (3^2)}{\log (2^2)}=\frac{\log (3)}{\log (2)}+\frac{2\log (3)}{2\log (2)}=2\frac{\log (3)}{\log (2)}$$ $$\log_2 (3)+\log_4 (9) =2\log_2 (3)$$
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 $log_4(9)+log_2(3)=log_2(3)+log_2(3)=2log_2(3)$
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 $\log_v 23^2=\log_v 529\neq \log_v 1127$
• This is a pet peeve of mine, but I think using parenthesis to surround the argument of logarithms (and functions like $\sin$, $\cos$, etc.) should become standard practice--this matches functional notations like $f(x)$ etc. and hopefully helps students understand the logarithm is a function (you'd be surprised how many times I had students think that $\log x$ meant $\log * x$--not that the parenthesis completely fixes that, but I think it helps)...just my opinion though. – Jared May 14 '15 at 23:47