Question 419982
First of all, there is something wrong with this problem:<ul><li>If {{{log(b, (16)) = 4}}} then {{{b^4 = 16}}}.</li><li>If {{{b^4 = 16}}} then b = 2 or -2.</li><li>Since b is the base of a logarithm it cannot be negative. So b must be 2.</li><li>If b = 2 then {{{log(b, (2)) = log(2, (2)) = 1}}} <i>not 3!</i></li><li>So the equations {{{log(b, (2)) = 3}}} and {{{log(b, (16)) = 4}}} are inconsistent. (IOW: It is not possible for both to be true at the same time.</li></ul>
If you posted the problem exactly as it was given to you, then your teacher (or the people who wrote the textbook) made a mistake. If you decide to bring this to your teacher's attention, the please do so respectfully.<br>
If we pretend that there there is no error in the problem itself, then the problem is to rewrite {{{log(b, (32))}}} in terms of the two logarithms your were given. Since 32 = 16*2 we can rewrite {{{log(b, (32))}}} as:
{{{log(b, (16*2))}}}
Now we can use a property of logarithms, {{{log(a, (p*q)) = log(a, (p)) + log(a, (q))}}}, which allows us to expression the log of a product as the sum of the logs of the factors:
{{{log(b, (16)) + log(b, (2))}}}
We have now expressed {{{log(b, (32))}}} in terms of the two logs you were given. We can replace these logs with the values we were given for them:
4 + 3
which simplifies to
7<br>