How do I express $\log_52$ in terms of $a$ and $b$ if:
$\log_62 =a$ and $\log_53 =b$
I've tried:
Converting the $a$ and $b$ equations to fractions, and substituting $\log2$ and $\log5$ with $a\log6$ and $(\log3)/b$ respectively, but I ended up with the same equation after simplifying things out.
I've also tried to convert the $2$ in $\log_52$ into fractions and go from there, but I went up going in circles and never got anywhere.
How would I solve this question?
Using change of base (change to base 5):
$$ \log_6 2 = \frac{\log_5 2}{\log_5 6} = \frac{\log_5 2}{\log_5 3 + \log_5 2}$$
$$ a = \frac{\log_5 2}{b + \log_5 2}$$
$$ ab = (1-a)\log_5 2$$
$$ \log_5 2 = ?$$
I trust you can finish the rest
$a=\log_62=\dfrac{\log2}{\log2+\log3}$ as $\log(ab)=\log a+\log b$ when all the logarithms are defined.
Now express $\log2$ in terms of $\log3$
Again, $b=\log_53=\dfrac{\log3}{\log5}\iff\log5=\dfrac{\log3}b$
Now $\log_52=\dfrac{\log2}{\log5}=?$
Alternatively,
$$\log_53=b\implies3=5^b$$
$$\log_62=a\implies2=6^a=(5^b)^a2^a\iff2^{1-a}=5^{ab}$$
Take log wrt base $5$ $$(1-a)\log_52=ab$$
so $a\log2+a\log3=\log2$,
$a\log2-\log2=a\log3$,
$a=-a\log3$
Is this correct?
– Marcel Yuwono Nov 01 '15 at 08:23