<div>
<a href="http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#log(double)" rel="ugc">http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#log(double)</a><br />
<a href="http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#E" rel="ugc">http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#E</a><br />
and <br />
<a href="http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#pow(double" rel="ugc">http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#pow(double</a>, double)<br />
<br />
have pretty good explanation of both log and pow (and how the result of the whole thing end up being the second parameter of the pow :)<br />
<br />
PS: d-glitch got it - I am just throwing a couple of links to help with the understanding. :)
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<div>
Thanks d-glitch for the explanation. I wasn't sure.<br />
<br />
Also thanks to Venablili!!!
</div>


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<div class="content wysiwyg-content">
I am having problems solving this math problem. I am doing this for fun. &nbsp;The answer is 13.0 but I don't understand I how they came up with the solution.<br />
<br />
7 - 2 + Math.log10(1000) + Math.log(Math.pow(Math.E, 5))<br />
<br />
I understand<br />
<br />
7 -2 + Math.log10(1000) = 8<br />
<br />
I not sure if about the second half of the problem.<br />
<br />
Math.log(Math.pow(Math.E, 5)<br />
<br />
Is the 5 that is used to raise E to the fifth power? The same 5 added to 8.<br />
<br />
If so can someone explain this to me?<br />
<br />
Thanks for your help.
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I am having problems solving this math problem. I am doing this for fun.  The answer is 13.0 but I don't understand I how they came up with the solution.

7 - 2 + Math.log10(1000) + Math.log(Math.pow(Math.E, 5))

I understand

7 -2 + Math.log10(1000) = 8

I not sure if about the second half of the problem.

Math.log(Math.pow(Math.E, 5)

Is the 5 that is used to raise E to the fifth power? The same 5 added to 8.

If so can someone explain this to me?

Thanks for your help.

http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#log(double)
http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#E
and 
http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#pow(double, double)

have pretty good explanation of both log and pow (and how the result of the whole thing end up being the second parameter of the pow :)

PS: d-glitch got it - I am just throwing a couple of links to help with the understanding. :)

Thanks d-glitch for the explanation. I wasn&amp;#39;t sure.

Also thanks to Venablili!!!

Java Math question

I am having problems solving this math problem. I am doing this for fun.  The answer is 13.0 but I don&amp;#39;t understand I how they came up with the solution.

7 - 2 + Math.log10(1000) + Math.log(Math.pow(Math.E, 5))

I understand

7 -2 + Math.log10(1000) = 8

I not sure if about the second half of the problem.

Math.log(Math.pow(Math.E, 5)

Is the 5 that is used to raise E to the fifth power? The same 5 added to 8.

If so can someone explain this to me?

Thanks for your help.

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