Question 718223
<ul><li>The expression,as you posted it, is {{{log((125^5))}}}. If this is correct, then just type it into your calculator.</li><li>If the expression is: {{{log(125, (5))}}} then...<ul><li>There is no exponent in this expression, even though it looks there is. The second 5 is <u>not</u> an exponent of the 125!! The 125 is the base of the logarithm and the 5 is the argument of the logarithm. Posting logarithms are difficult. To post a log like this, either<ul><li>Use some English to help describe the log. For example, this one could be described as: "the base 125 log of 5"; or...</li><li>Teach yourself the syntax we use on algebra.com to get nice-looking logs like {{{log(2, (15))}}}. Click on the "Show source" link above to see what I've typed to get these logs to look good.</li></ul></li><li>The expression {{{log(125, (5))}}} represents the exponent one would put on 125 to get a result of 5. So this problem is asking us to figure out: What power of 125 is 5? Perhaps you already know the answer. Perhaps knowing that {{{5^3 = 125}}} would help you figure out the answer. (It's not 3. 3 is the exponent we put on 5 to get 125. We're looking for the exponent to put on 125 to get 5.) If you still don't know the answer, then we can use {{{5^3 = 125}}} to figure it out. We will use the change of base formula, {{{log(a, (p)) = log(b, (p))/log(b, (a))}}} to convert the base 125 log to an expression of base 5 logs:
{{{log(125, (5)) = log(5, (5))/log(5, (125))}}}
The numerator of this fraction is the exponent you put on a 5 to get 5. This should be 1, obviously. The denominator is the exponent you put on a 5 to get 125. From {{{5^3 = 125}}} we know that this exponent is 3. So now we have:
{{{log(125, (5)) = log(5, (5))/log(5, (125)) = 1/3}}}
I hope this answer makes sense now. An exponent of 1/3, remember, means cube root.</li></ul>