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Beefdawg
07-24-2011, 03:44 PM
Hello all, I have a final exam tomorrow and was wondering if I can get some help with a certain problem:

What is log base 8 of radical 2 over 32? To make it more clear, its radical 2 over just the number 32. 32 is not in a radical. Thanks for the help.

Subhotosh Khan
07-24-2011, 06:28 PM
Hello all, I have a final exam tomorrow and was wondering if I can get some help with a certain problem:

What is log base 8 of radical 2 over 32? To make it more clear, its radical 2 over just the number 32. 32 is not in a radical. Thanks for the help.

Start with definition of log:

Log[sub:2oaimvei]a[/sub:2oaimvei]b = x ................? ...............a[sup:2oaimvei]x[/sup:2oaimvei] = b

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galactus
07-24-2011, 06:45 PM
Try the change of base formula:

log_{a}b=\frac{log(b)}{log(a)}

This means we have \frac{log(\frac{\sqrt{2}}{32})}{log(8)}

Now, use the two rules log(a/b)=log(a)-log(b) and log(a^{b})=b\cdot log(a) to write it in terms of log(2).

It will whittle clean down to a little ol' fraction.

JeffM
07-24-2011, 07:08 PM
Hello all, I have a final exam tomorrow and was wondering if I can get some help with a certain problem:

What is log base 8 of radical 2 over 32? To make it more clear, its radical 2 over just the number 32. 32 is not in a radical. Thanks for the help.
OK Galactus beat me to it again. Another way to analyze it is to see that everything is a power of 2: radical 2 = 2[sup:2l33wd49]1/2[/sup:2l33wd49], 8 = 2[sup:2l33wd49]3[/sup:2l33wd49], and 32 = 2[sup:2l33wd49]5[/sup:2l33wd49]. So everything is also a power of 8: radical 2 = 8[sup:2l33wd49]1/6[/sup:2l33wd49], 8 = 8[sup:2l33wd49]1[/sup:2l33wd49], and 32 = 8[sup:2l33wd49]5/3[/sup:2l33wd49] = 8[sup:2l33wd49]10/6[/sup:2l33wd49].
So log[sub:2l33wd49]8[/sub:2l33wd49][(8[sup:2l33wd49]1/6[/sup:2l33wd49])/(8[sup:2l33wd49]10/6[/sup:2l33wd49])] = ????