Question 594466
Given:
{{{log(a, (5))=2.3}}}
{{{log(a, (3))=1.6}}}
Plus,
{{{log(a, (a)) = 1}}} (This is always true, no matter what "a" is, so it does not need to be told this.)
you have been asked to find various other base a logarithms. The "trick" to these is to use algebra and/or properties of logarithms to rewrite the desired logs in terms of the logs you already know.<br>
So for
{{{log(a, (15))}}}
we want to rewrite the 15 in terms of 5's, 3's and/or a's. I hope that you can see that 15 is 5*3. Replacing the 15 with 5*3 we get:
{{{log(a, (5*3))}}}
Now we use a property of logarithms for logs of a product, {{{log(x, (p*q)) = log(x, (p)) + log(x, (q))}}}, we can separate the 5 and 3:
{{{log(a, (5)) + log(a, (3))}}}
Now that we have the log of 15 expressed in terms of logs of 5 and 3. We can now use the given values:
2.3 + 1.6
which simplifies to 3.9. So {{{log(a, (15)) = 3.9}}}<br>
{{{log(a, (9))}}}
For 9 we could use either 3*3 or {{{3^2}}}. I'll use the later one so you can see another property in use:
{{{log(a, (3^2))}}}
Using a property for logs of a power, {{{log(x, (p^q)) = q*log(x, (q))}}} we can separate the exponent from the 3:
{{{2*log(a, (3))}}}
Replacing the log with its given value we get:
2 * 1.6
which simplifies to
3.2
So {{{log(a, (9)) = 3.2}}}<br>
{{{log(a, (5/3))}}}
5/3 is already expressed in terms of 5's and 3's. Using a property for logs of quotients, {{{log(x, (p/q)) = log(x, (p)) - log(x, (q))}}} we get:
{{{log(a, (5)) - log(a, (3))}}}
Replacing the logs with their given values we get:
2.3 - 1.6
which simplifies to
0.7
So {{{log(a, (5/3)) = 0.7}}}<br>
{{{log(a, (3/a^2))}}}
First we'll use the property for quotients:
{{{log(a, (3)) - log(a, (a^2))}}}
and then the property for powers (on the second log):
{{{log(a, (3)) - 2*log(a, (a))}}}
Now we can replace the logs with their known values.
1.6 - 2*1
which simplifies
1.6 - 2
-0.4
So {{{log(a, (3/a^2)) = -0.4}}}