Question 618078
{{{ log( 2, (3)) - 2log( 2, (4)) + 3log ( 2, (3))}}}
I assume the problem is to rewrite this expression as a single logarithm. (Please include the problem's directions in the future.)<br>
The first and last logs are like terms (same base, same argument). So we can add them:
{{{ 4log( 2, (3)) - 2log( 2, (4)) }}}<br>
The remaining terms are not like terms. But there are some properties of logarithms that provide another way to combine terms:<ul><li>{{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}}</li><li>{{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}}</li></ul>These properties require logs of the same base ... with coefficients of 1. Our logs have the same base but they do not have coefficients of 1.<br>
Fortunately there is yet another property of logarithms, {{{q*log(a, (p)) = log(a, (p^q))}}}, which allows us to "move" a coefficient into the argument as its exponent. Using this property on our logs:
{{{ log( 2, (3^4)) - log( 2, (4^2)) }}}
which simplifies to:
{{{ log( 2, (81)) - log( 2, 16) }}}<br>
Now we can use one of the earlier properties to combine the two logs. We use the second one because its logs, like ours, have a "-" between them:
{{{ log( 2, (81/16)) }}}
The fraction does not reduce so, unless you want a mixed number or decimal for 81/16, we are finished.