Question 270469
{{{log(6, (27))}}}
The key to this problem is to realize that a base 6 logarithm is hard to work with. We need to change the base to one that is easy to work with, like base 10 logarithms. Fortunately there is a formula for doing just this: {{{log(a, (x)) = log(b, (x))/log(b, (a))}}}. Using this formula to change your base 6 logarithm to an expression of base 10 logarithms we get:
{{{log(6, (27)) = log((27))/log((6))}}}
We now have base 10 logarithms to work with. Next, since we are given log(2) and log(3), we want to express the arguments in terms of 2's and/or 3's. You already figured out how to express a 6 in terms of 2's and 3's. With a little thought I hope you see that {{{27 = 3^3}}}. So rewriting the logarithms using 2's and 3's we get:
{{{log(6, (27)) = log((27))/log((6)) = log((3^3))/log((2*3))}}}
Next we can use one property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}} to move the exponent of {{{3^3}}} out in front of the logarithm. And we can use another property of logarithms, {{{log(a, (p*q)) = log(a, (p)) + log(a, (b))}}}, to split the log(2*3) into separate logarithms:
{{{log(6, (27)) = log((27))/log((6)) = log((3^3))/log((2*3)) = (3*log((3)))/(log((2)) + log(3)))}}}
We have finally changed our original base 6 logarithm of 27 into an expression of base 10 logarithms of 2 and 3. We can now use the given values for log(2) and log(3) and simplify:
{{{log(6, (27)) = log((27))/log((6)) = log((3^3))/log((2*3)) = (3*log((3)))/(log((2)) + log((3))) = (3*(0.477))/(0.301 + 0.477) = 1.431/0.778 = 1.8393316195372751}}}
Rounded to the nearest 1/100 this matches the answer you got. But I have no idea how you came up with this. The work you provided is grossly incomplete.