```Question 425890
{{{7^log(7, (45)) + log(6, (3600))}}}
Once you understand logarithms you will find the first term ridiculously easy to simplify.<br>
Generally logarithms are exponents. {{{log(a, (b))}}} represents the exponent for "a" that results in "b". In other "words", {{{a^log(a, (b)) = b}}} <i>by definition of what a logarithm is!</i>.<br>
So {{{7^log(7, (45)) = 45}}} by definition. Now your expression is:
{{{45 + log(6, (3600))}}}
This is a single logarithm expression so it may be an acceptable answer. But the second logarithm can be simplified, too, because its argument, 3600, has a power of 6 factor:
{{{45 + log(6, (36*100))}}}
{{{45 + log(6, (36)) +log(6, (100))}}}
{{{45 + log(6, (6^2)) +log(6, (100))}}}
{{{45 + 2 +log(6, (100))}}}
{{{47 +log(6, (100))}}}
This is a simplified, single logarithm expression.<br>
If the problem is literally to write the expression as just a single logarithm, then we need both terms to be in terms of base 6 logarithms. Since {{{log(6, (6)) = 1}}} and since multiplying by a 1 (no matter what form it is in) does not change a number we can rewrite the expression as:
{{{47*log(6, (6)) +log(6, (100))}}}
Now we can use a property of logarithms, {{{q*log(a, (p)) = log(a, (p^q))}}} to move the coefficient of the first logarithm in the argumetn as its exponent:
{{{log(6, (6^47)) +log(6, (100))}}}
Now we can use another property of logarithms, {{{log(a, (p)) + log(a, (q))}}}, to combine the two logarithms into one. (This property requires that the two logarithms have the same base (which is why we introduced the base 6 log to the first term) and coefficients of 1 (which is why we used the other property to move the 47 into the argument)) Using this property to combine the two logarithms we get:
{{{log(6, (6^47*100))}}}
or
{{{log(6, (100*6^47))}}}```