Question 719537
These problems where you're asked to express one logarithm in terms of a certain set of given logs usually involves finding a way to express the log in terms of logs of products, quotients or powers of the arguments of the given logs. In this problem we will try to express log(127/27) in terms of logs (of any base) of products, quotients and/or powers of 2's, 3's and/or 5's.<br>
What is often overlooked in these types of problems is: logs of powers of the base of that log. For example, if we were trying to express something in terms of base 10 logs of 2, 3 or 5, we can also use logs of powers of 10. We can do this because base 10 logs of powers of 10 simplify to a simple number.<br>
When looking at log(127/27) we might see that the 27 could be expressed as a power of 3: {{{27 = 3^3}}}. But what about the 127? 127 is prime. It is not a product, quotient and/or power of 2's, 3's and/or 5's. The only way we can handle the 127 is: To use base 127 logs!<br>
So we will start by using the base conversion formula, {{{log(a, (p)) = log(b, (p))/log(b, a))}}} to convert the log(127/27) to base 127 logs:
{{{log(127, (127/27))/log(127, 10))}}}<br>
Now we can use a property of logarithms, {{{log(a, (p/q)) = log(a, (p)) - log(a, (q))}}}, to rewrite the log in the numerator:
{{{(log(127, (127)) - log(127, (27)))/log(127, (10))}}}
The first log is the base 127 log of a known power of 127, i.e. {{{127^1}}}. So that log is a 1:
{{{(1 - log(127, (27)))/log(127, (10))}}}
For the remaining logs we can express the numerator's argument as a power of 3 and the denominator's argument as a product of 2 and 5:
{{{(1 - log(127, (3^3)))/log(127, (2*5))}}}
For the log in the numerator we can use another property of logarithms, {{{log(a, (p^n)) = n*log(a,(p))}}}. For the log in the denominator we can use a third property logs: {{{log(a, (p*q)) = log(a, (p)) + log(a, (q))}}}.
{{{(1 - 3*log(127, (3)))/(log(127, (2))+log(127, (5)))}}}
We have now expressed
{{{log((127/27))}}}
in terms of base 127 logs of 2's, 3's and 5's:
{{{(1 - 3*log(127, (3)))/(log(127, (2))+log(127, (5)))}}}