Question 401308
{{{log(12, (3)) = 0.4421}}}
{{{log(12, (7)) = 0.7831}}}
We are given these two logarithms. Another base 12 logarithm we can use without having to be told is:
{{{log(12, (12)) = 1}}}<br>
The "trick" to this kind of problem is to express the argument of the desired logarithm in terms of products, quotients and/or powers of the arguments of the known logarithms. In this problem the argument of the desired logarithm is 48. The arguments of the known logarithms are 3, 7 and 12. So we want to express 48 as some product(s), quotient(s) and/or power(s) of 3, 7 and/or 12.<br>
First we can find that 48=12*4. But we don't know what {{{log(12, (4))}}} is. Can we express a 4 in terms of 3, 7 and/or 12? Answer: 4 = 12/3. So now we have
{{{48 = 12*(12/3)}}}<br>
Now we can find {{{log(12, (48))}}}:
{{{log(12, (48)) = log(12, (12*(12/3)))}}}
Now we can use a property of logarithms, {{{log(a, (p*q)) = log(a, (p)) + log(a, (q))}}}, to split the logarithm of a product into the sum of the logarithms of the factors:
{{{log(12, (48)) = log(12, (12*(12/3))) = log(12, (12)) + log(12, (12/3))}}}
Next we can use another property of logarithms, {{{log(a, (p/q)) = log(a, (p)) - log(a, (q))}}}, to split the logarithm of a quotient into the difference of the logarithms of the numerator and denominator:
{{{log(12, (48)) = log(12, (12*(12/3))) = log(12, (12)) + log(12, (12/3)) = log(12, (12)) + (log(12, (12)) - log(12, (3)))}}}
We can now replace each logarithm with its known value:
{{{log(12, (48)) = log(12, (12*(12/3))) = log(12, (12)) + log(12, (12/3)) = log(12, (12)) + (log(12, (12)) - log(12, (3))) = 1 + (1 - 0.4421)}}}
And simplify:
{{{log(12, (48)) = log(12, (12*(12/3))) = log(12, (12)) + log(12, (12/3)) = log(12, (12)) + (log(12, (12)) - log(12, (3))) = 1 + (1 - 0.4421) = 1.5579}}}